14 Large sample test
A sample with a size of 30 or more (\(n≥30\)) is generally referred to as a large sample. Statistical methods based on such samples are known as large-sample tests. These methods rely on large-sample theory, which uses the Central Limit Theorem to approximate the sampling distribution of test statistics by the normal distribution, even when the population distribution is unknown.
14.1 Central Limit Theorem
The Central Limit Theorem states that when the sample size is sufficiently large (generally \(n≥30\)), the sampling distribution of the sample mean is approximately normal, irrespective of the shape of the population distribution, provided the observations are independent and the population has a finite variance.
Mathematically, if a random sample of size \(n\) is drawn from a population with mean \(\mu\) and variance \(\sigma^2\), then as \(n\) \(\rightarrow\) \(\infty\),
\[ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \xrightarrow{d} N(0,1), \tag{14.1}\]
where:
- \(\bar{X}\) = sample mean
- \(\mu\) = population mean
- \(\sigma\) = population standard deviation
- \(N(0,1)\) = standard normal distribution
- \(\xrightarrow{d}\) denotes convergence in distribution. For now just think as “tends to follow” or “approaches” the standard normal distribution as the sample size increases.
14.2 Z test
The \(Z\)-test is a statistical test based on the standard normal distribution. It is used to test hypotheses about population means and proportions when the sample size is sufficiently large and the necessary assumptions are satisfied. The value of the test statistic is compared with the critical value from the standard normal distribution to decide whether the null hypothesis should be accepted or rejected.
The test statistic \(Z\) is calculated from a large sample using appropriate formulas that we will discuss in the coming sections. \(Z\) follows a standard normal distribution, \(Z \sim N(0,1)\). The following large sample tests are covered in this chapter:
Test for a single proportion
Test for the equality of two proportions
Test for a single mean
Test for the equality of two means
14.2.1 Decision rule for the Z test
Let \(Z\) be the calculated value and \(\alpha\) be the level of significance. We reject the null hypothesis if:
\(|Z| > Z_{\alpha/2}\) ; for a two-tailed test
\(Z > Z_{\alpha}\) ; for a right-tailed test
\(Z < -Z_{\alpha}\) ; for a left-tailed test
The critical values of \(Z\) for specified levels of significance are given in Table 14.1.
| \(\alpha\) (one tail) | \(Z\) value |
|---|---|
| 0.10 | 1.282 |
| 0.05 | 1.645 |
| 0.025 | 1.960 |
| 0.01 | 2.326 |
| 0.005 | 2.576 |
For a two-tailed test, the significance level \(\alpha\) is split equally between the two tails, so the critical value corresponds to the one-tail value at \(\alpha/2\). Table 14.2 gives the critical values directly in terms of the overall two-tailed \(\alpha\).
| \(\alpha\) (two tail) | \(Z\) value |
|---|---|
| 0.20 | 1.282 |
| 0.10 | 1.645 |
| 0.05 | 1.960 |
| 0.02 | 2.326 |
| 0.01 | 2.576 |
For a two-tailed test at \(\alpha\) = 0.05, we split the significance level equally between the two tails and use the value at \(\alpha/2\) = 0.025, which is \(Z_{0.025}\) = 1.960. Similarly, for a two-tailed test at \(\alpha\) = 0.01 we use \(Z_{0.005}\) = 2.576. These are the same critical values shown directly against the overall two-tailed \(\alpha\) in Table 14.2.
14.3 Test for a single population proportion
Consider a population with proportion \(P\), where \(P\) is unknown. We take a random sample of size \(n\) from the population and calculate the sample proportion \(p\). We want to test whether the population proportion \(P\) is equal to a specified value \(P_0\), based on the sample proportion \(p\). Note here we always frame null hypothesis on population parameter and test using the sample statistics.
The null hypothesis to be tested is:
\(H_0\): \(P\) = \(P_0\)
The alternative hypothesis may be either:
\(H_1\): \(P\) < \(P_0\) (left-tailed alternative)
\(H_1\): \(P\) > \(P_0\) (right-tailed alternative)
\(H_1\): \(P\) \(\mathbf{\neq}\) \(P_0\) (two-tailed alternative)
For this test, we calculate the test statistic \(Z\) using the following formula:
\[Z = \frac{p - P_0}{\sqrt{P_0 Q_0 / n}} \tag{14.2}\]
where \(Q_0 = 1 - P_0\).
When the null hypothesis is true, this \(Z\) follows a standard normal distribution with mean 0 and variance 1, \(Z \sim N(0,1)\).
The calculated value of \(Z\) is then compared with the critical value of \(Z\) from the table of the standard normal distribution. If the calculated value is higher than the table value, we reject the null hypothesis. If the calculated value is less than the table value, we conclude that we do not have enough evidence in our sample to reject the null hypothesis. See Section 14.2.1.
The standard error in the single-proportion test is computed using the null value \(P_0\), giving \(\sqrt{P_0 Q_0/n}\), because the test statistic is derived under the assumption that the null hypothesis is true. Some textbooks use the sample-based standard error \(\sqrt{pq/n}\); for large samples the two give very similar results.
Example 14.1 A sample of 500 apples was taken from a large consignment and 60 were found to be bad. Test whether the proportion of bad apples in the consignment is 0.1 or not, at the 5% level of significance.
Solution
Sample size (\(n\)) = 500
Proportion of bad apples in the sample (\(p\)) = 60/500 = 0.12
We want to test whether the proportion of bad apples in the population is 0.1, so \(P_0\) = 0.1 and \(Q_0\) = 1 - 0.1 = 0.9.
The null hypothesis is:
\(H_0\): \(P\) = 0.1
Since we want to test whether the proportion is 0.1 or not, we use a two-tailed test with the alternative hypothesis:
\(H_1\): \(P\) \(\mathbf{\neq}\) 0.1
The level of significance is 5%, so \(\alpha\) = 0.05.
Now we calculate \(Z\) using Equation 14.2:
\[Z = \frac{p - P_0}{\sqrt{P_0 Q_0 / n}} = \frac{0.12 - 0.1}{\sqrt{0.1 \times 0.9 / 500}}\]
\[= \frac{0.02}{0.01342} = 1.49\]
Since it is a two-tailed test, we compare with the critical value at \(\alpha/2\) = 0.025, which is 1.960 (see Section 14.2.1). The calculated value (1.49) is less than the table value (1.960), so we do not have enough evidence to reject the null hypothesis. We conclude that there is insufficient evidence to say the proportion of bad apples differs from 0.1.
Exercise 14.1 A random sample of 500 plants was taken from a large experimental field and 65 plants were found to be affected by yellowing disease. Test whether the proportion of affected plants differs significantly from 0.1. (\(\alpha\) = 0.05)
14.4 Test for equality of two proportions
Consider two populations with proportions \(P_1\) and \(P_2\), both unknown. We take random samples of sizes \(n_1\) and \(n_2\) from the two populations respectively and calculate the sample proportions \(p_1\) and \(p_2\). We want to test whether the population proportions \(P_1\) and \(P_2\) are equal, based on the sample proportions.
The null hypothesis to be tested is:
\(H_0\): \(P_1\) = \(P_2\)
The alternative hypothesis may be either:
\(H_1\): \(P_1\) < \(P_2\) (left-tailed alternative)
\(H_1\): \(P_1\) > \(P_2\) (right-tailed alternative)
\(H_1\): \(P_1\) \(\mathbf{\neq}\) \(P_2\) (two-tailed alternative)
For this test, we calculate the test statistic \(Z\) using the following formula:
\[Z = \frac{p_1 - p_2}{\sqrt{\widehat{P}\,\widehat{Q}\left( \dfrac{1}{n_1} + \dfrac{1}{n_2} \right)}} \tag{14.3}\]
where \(\widehat{P} = \dfrac{n_1 p_1 + n_2 p_2}{n_1 + n_2}\) is the pooled proportion and \(\widehat{Q} = 1 - \widehat{P}\).
When the null hypothesis is true, this \(Z\) follows a standard normal distribution, \(Z \sim N(0,1)\).
The calculated value of \(Z\) is compared with the critical value of \(Z\) from the standard normal table (see Section 14.2.1). If the calculated value is higher than the table value, we reject the null hypothesis. Otherwise, we conclude that we do not have enough evidence to reject the null hypothesis.
Example 14.2 In order to assess the adoption of a new variety of paddy by farmers, a survey was conducted in a locality. The survey covered 80 farmers with large land holdings and 250 farmers with small land holdings. It was observed that 50 of the large-holding farmers and 78 of the small-holding farmers adopted the new paddy variety. Test whether there is any significant difference in the adoption behaviour of the two groups of farmers. (\(\alpha\) = 0.01)
Solution
Sample size of large-holding farmers (\(n_1\)) = 80
Sample size of small-holding farmers (\(n_2\)) = 250
Proportion of large-holding farmers who adopted the variety (\(p_1\)) = 50/80 = 0.625
Proportion of small-holding farmers who adopted the variety (\(p_2\)) = 78/250 = 0.312
We want to test whether the proportion is significantly different between the two populations, so:
\(H_0\): \(P_1\) = \(P_2\)
\(H_1\): \(P_1\) \(\mathbf{\neq}\) \(P_2\)
This is a two-tailed test.
Step 1: Calculate the pooled proportion.
\[\widehat{P} = \frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac{80 \times 0.625 + 250 \times 0.312}{80 + 250} = 0.3879\]
\[\widehat{Q} = 1 - 0.3879 = 0.6121\]
Step 2: The level of significance is 1%, so \(\alpha\) = 0.01.
Step 3: Calculate \(Z\) using Equation 14.3.
\[Z = \frac{p_1 - p_2}{\sqrt{\widehat{P}\,\widehat{Q}\left( \dfrac{1}{n_1} + \dfrac{1}{n_2} \right)}}\]
\[= \frac{0.625 - 0.312}{\sqrt{0.3879 \times 0.6121 \left( \dfrac{1}{80} + \dfrac{1}{250} \right)}}\]
\[= \frac{0.313}{\sqrt{0.2374 \times 0.0165}} = \frac{0.313}{0.06258} = 5.00\]
Since it is a two-tailed test, we compare with the critical value at \(\alpha/2\) = 0.005, which is 2.576 (see Section 14.2.1). Since the calculated value (5.00) is greater than the table value (2.576), we reject the null hypothesis and conclude that there is a significant difference between the two population proportions.
14.5 Test for a single population mean
Consider a population with mean \(\mu\), where \(\mu\) is unknown. We take a random sample of size \(n\) from the population and calculate the sample mean \(\overline{x}\). We want to test whether the population mean \(\mu\) is equal to a known constant \(\mu_0\), based on the sample mean.
The null hypothesis to be tested is:
\(H_0\): \(\mu\) = \(\mu_0\)
The alternative hypothesis may be either:
\(H_1\): \(\mu\) < \(\mu_0\) (left-tailed alternative)
\(H_1\): \(\mu\) > \(\mu_0\) (right-tailed alternative)
\(H_1\): \(\mu\) \(\mathbf{\neq}\) \(\mu_0\) (two-tailed alternative)
This test has two cases:
Case when the population standard deviation \(\sigma\) is known
Case when the population standard deviation \(\sigma\) is unknown
14.5.1 Standard deviation is known
When \(\sigma\) is known, we calculate the test statistic \(Z\) using the following formula:
\[Z = \frac{\overline{x} - \mu_0}{\dfrac{\sigma}{\sqrt{n}}} \tag{14.4}\]
When the null hypothesis is true, this \(Z\) follows a standard normal distribution, \(Z \sim N(0,1)\). The decision rule is given in Section 14.2.1.
Example 14.3 A fertilizer packing unit is set to fill bags with a mean weight of 50 kg. From long experience the population standard deviation of the filling process is known to be 3.2 kg. A random sample of 64 bags was drawn and the mean weight was found to be 49.2 kg. Test whether the packing process is working at the intended mean weight of 50 kg, at the 5% level of significance.
Solution
Null hypothesis, \(H_0\): \(\mu\) = 50
Alternative hypothesis, \(H_1\): \(\mu\) \(\mathbf{\neq}\) 50 (two-tailed test)
Sample size (\(n\)) = 64
Sample mean (\(\overline{x}\)) = 49.2
Population standard deviation (\(\sigma\)) = 3.2
Population mean (\(\mu_0\)) = 50
Level of significance, \(\alpha\) = 0.05
Since \(\sigma\) is known, we use Equation 14.4:
\[Z = \frac{\overline{x} - \mu_0}{\dfrac{\sigma}{\sqrt{n}}} = \frac{49.2 - 50}{\dfrac{3.2}{\sqrt{64}}}\]
\[= \frac{-0.8}{0.4} = -2.0\]
Since it is a two-tailed test, we compare \(|Z|\) with the critical value at \(\alpha/2\) = 0.025, which is 1.960 (see Section 14.2.1). Since the calculated value (2.0) is greater than the table value (1.960), we reject the null hypothesis. We conclude that the packing process is not working at the intended mean weight of 50 kg.
14.5.2 Standard deviation is unknown
When \(\sigma\) is unknown, it is replaced by its sample estimate \(s\), and we calculate the test statistic \(Z\) using the following formula:
\[Z = \frac{\overline{x} - \mu_0}{\dfrac{s}{\sqrt{n}}} \tag{14.5}\]
When the null hypothesis is true, this \(Z\) follows a standard normal distribution, \(Z \sim N(0,1)\). The decision rule is given in Section 14.2.1.
Example 14.4 A sample of 900 items has a mean of 3.4 cm and a standard deviation of 2.61 cm. Test whether the population mean is 3.25 cm at the 5% significance level.
Solution
Null hypothesis, \(H_0\): \(\mu\) = 3.25
Alternative hypothesis, \(H_1\): \(\mu\) \(\mathbf{\neq}\) 3.25 (two-tailed test)
Sample size (\(n\)) = 900
Sample mean (\(\overline{x}\)) = 3.4
Sample standard deviation (\(s\)) = 2.61
Population mean (\(\mu_0\)) = 3.25
Level of significance, \(\alpha\) = 0.05
Since \(\sigma\) is unknown, we use \(s\) in place of \(\sigma\), applying Equation 14.5:
\[Z = \frac{\overline{x} - \mu_0}{\dfrac{s}{\sqrt{n}}} = \frac{3.4 - 3.25}{\dfrac{2.61}{\sqrt{900}}}\]
\[= \frac{0.15}{0.087} = 1.724\]
Since it is a two-tailed test, we compare with the critical value at \(\alpha/2\) = 0.025, which is 1.960 (see Section 14.2.1). Since the calculated value (1.724) is less than the table value (1.960), we do not have enough evidence to reject the null hypothesis. We conclude that there is insufficient evidence to say the mean differs from 3.25 cm.
Exercise 14.2 From a paddy field a sample of 36 plants was selected at random and the panicle lengths were observed. The mean and standard deviation of these measurements were 18.7 cm and 1.25 cm respectively. Test whether the mean panicle length of paddy is 19 cm. (\(\alpha\) = 0.05)
14.6 Test for equality of two means
Let there be two normally distributed populations with means \(\mu_1\) and \(\mu_2\) and standard deviations \(\sigma_1\) and \(\sigma_2\) respectively. Let samples of sizes \(n_1\) and \(n_2\) be taken from these populations, with sample means \(\overline{x}_1\) and \(\overline{x}_2\). We want to test whether the two population means are significantly different, based on the sample means.
The null hypothesis to be tested is:
\(H_0\): \(\mu_1\) = \(\mu_2\)
The alternative hypothesis may be either:
\(H_1\): \(\mu_1\) < \(\mu_2\) (left-tailed alternative)
\(H_1\): \(\mu_1\) > \(\mu_2\) (right-tailed alternative)
\(H_1\): \(\mu_1\) \(\mathbf{\neq}\) \(\mu_2\) (two-tailed alternative)
There are two cases under this test:
Case when the population standard deviations are equal, \(\sigma_1 = \sigma_2 = \sigma\)
Case when the population standard deviations are not equal, \(\sigma_1 \neq \sigma_2\)
14.6.1 Standard deviations are equal
When the population standard deviations are equal, we calculate the test statistic \(Z\) using the following formula:
\[Z = \frac{\overline{x}_1 - \overline{x}_2}{\sigma \sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}} \tag{14.6}\]
When the null hypothesis is true, this \(Z\) follows a standard normal distribution, \(Z \sim N(0,1)\). The decision rule is the same as above (see Section 14.2.1).
If the common \(\sigma\) is unknown, it is replaced by its pooled estimate \(\sqrt{\dfrac{n_1 s_1^2 + n_2 s_2^2}{n_1 + n_2}}\), where \(n_1\) and \(n_2\) are the sample sizes and \(s_1\) and \(s_2\) are the sample standard deviations.
Example 14.5 The means of two large samples of sizes 1000 and 2000 members are 67.5 inches and 68 inches respectively. Can the samples be regarded as drawn from the same population with standard deviation 2.5 inches? (Test at the 5% significance level.)
Solution
Null hypothesis, \(H_0\): \(\mu_1\) = \(\mu_2\)
Alternative hypothesis, \(H_1\): \(\mu_1\) \(\mathbf{\neq}\) \(\mu_2\) (two-tailed test)
Sample size (\(n_1\)) = 2000
Sample size (\(n_2\)) = 1000
Sample mean of the first group (\(\overline{x}_1\)) = 68
Sample mean of the second group (\(\overline{x}_2\)) = 67.5
Population standard deviation (\(\sigma\)) = 2.5
Level of significance, \(\alpha\) = 0.05
Since the two samples are assumed to come from the same population, the standard deviations are equal. We calculate \(Z\) using Equation 14.6:
\[Z = \frac{\overline{x}_1 - \overline{x}_2}{\sigma \sqrt{\dfrac{1}{n_1} + \dfrac{1}{n_2}}}\]
\[= \frac{68 - 67.5}{2.5 \sqrt{\dfrac{1}{2000} + \dfrac{1}{1000}}}\]
\[= \frac{0.5}{0.0968} = 5.16\]
Since it is a two-tailed test, we compare with the critical value at \(\alpha/2\) = 0.025, which is 1.960 (see Section 14.2.1). Since the calculated value (5.16) is greater than the table value (1.960), we reject the null hypothesis. We conclude that the samples cannot be regarded as drawn from the same population.
14.6.2 Standard deviations are not equal
When the population standard deviations are not equal, we calculate the test statistic \(Z\) using the following formula:
\[Z = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}} \tag{14.7}\]
When the null hypothesis is true, this \(Z\) follows a standard normal distribution, \(Z \sim N(0,1)\). The decision rule is the same as above (see Section 14.2.1).
Example 14.6 Two paddy varieties were compared for their grain yield. A random sample of 100 plots of variety A gave a mean yield of 52 quintals per hectare with a population standard deviation of 8 quintals per hectare, while a random sample of 100 plots of variety B gave a mean yield of 49 quintals per hectare with a population standard deviation of 6 quintals per hectare. The two population standard deviations are known and unequal. Test whether the two varieties differ significantly in mean yield, at the 5% level of significance.
Solution
Null hypothesis, \(H_0\): \(\mu_1\) = \(\mu_2\)
Alternative hypothesis, \(H_1\): \(\mu_1\) \(\mathbf{\neq}\) \(\mu_2\) (two-tailed test)
Sample size (\(n_1\)) = 100
Sample size (\(n_2\)) = 100
Sample mean of variety A (\(\overline{x}_1\)) = 52
Sample mean of variety B (\(\overline{x}_2\)) = 49
Population standard deviation of variety A (\(\sigma_1\)) = 8
Population standard deviation of variety B (\(\sigma_2\)) = 6
Level of significance, \(\alpha\) = 0.05
Since the two population standard deviations are unequal, we use Equation 14.7:
\[Z = \frac{\overline{x}_1 - \overline{x}_2}{\sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma_2^2}{n_2}}}\]
\[= \frac{52 - 49}{\sqrt{\dfrac{8^2}{100} + \dfrac{6^2}{100}}}\]
\[= \frac{3}{\sqrt{0.64 + 0.36}} = \frac{3}{1} = 3.0\]
Since it is a two-tailed test, we compare \(|Z|\) with the critical value at \(\alpha/2\) = 0.025, which is 1.960 (see Section 14.2.1). Since the calculated value (3.0) is greater than the table value (1.960), we reject the null hypothesis. We conclude that the two paddy varieties differ significantly in mean yield.
Exercise 14.3 Two random samples were drawn from two populations and the following data were obtained. Test whether the population means are equal. \(n_1\) = 400, \(n_2\) = 400, \(\overline{x}_1\) = 250, \(\overline{x}_2\) = 220, \(s_1\) = 40, \(s_2\) = 55. (\(\alpha\) = 0.05)
14.7 Chapter Summary
Fill in the blanks
Answers are given at the end of the chapter.
A sample with a size of __________ or more is generally referred to as a large sample.
Large-sample tests are based mainly on the __________ Limit Theorem.
According to the Central Limit Theorem, the sampling distribution of the sample mean approaches a __________ distribution as the sample size increases.
The standard normal distribution has mean __________ and variance __________.
The test based on the standard normal distribution is called the __________ test.
The test statistic used in a large-sample Z-test follows the __________ normal distribution when \(H_0\) is true.
The four large-sample tests discussed in the chapter are tests for a single proportion, equality of two proportions, a single mean, and equality of two __________.
In a test for a single population proportion, the null hypothesis is \(H_0:P=\) __________.
In the single-proportion test, \(Q_0=\) __________.
The standard error for testing a single population proportion is calculated using the __________ value of the proportion.
In the test for equality of two proportions, the pooled proportion is denoted by __________.
In the test for equality of two proportions, the pooled probability of failure is \(\widehat{Q}=\) __________.
In a test for a single population mean, the null hypothesis is \(H_0:\mu=\) __________.
When the population standard deviation is known, the test statistic for a single mean uses __________.
When the population standard deviation is unknown, the population standard deviation is replaced by the __________ standard deviation.
In the test for equality of two means, the null hypothesis is \(H_0:\mu_1=\) __________.
When two population standard deviations are equal, the common standard deviation is denoted by __________.
When two population standard deviations are unequal, the standard error contains __________ separate variance terms.
For a two-tailed Z-test at \(\alpha=0.05\), the critical value is __________.
For a two-tailed Z-test at \(\alpha=0.01\), the critical value is __________.
For a right-tailed Z-test at \(\alpha=0.05\), the critical value is __________.
For a left-tailed Z-test at \(\alpha=0.05\), the critical value is __________.
For a two-tailed test, the significance level \(\alpha\) is divided equally between the __________ tails.
For a two-tailed test, \(H_0\) is rejected when \(|Z|\) is __________ than the critical value.
For a right-tailed test, \(H_0\) is rejected when \(Z\) is __________ than the critical value.
For a left-tailed test, \(H_0\) is rejected when \(Z\) is __________ than the negative critical value.
The sample proportion is denoted by __________.
The sample mean is denoted by __________.
The population proportion is denoted by __________.
The population mean is denoted by __________.
The population standard deviation is denoted by __________.
The sample standard deviation is denoted by __________.
The difference between the sample mean and the hypothesized population mean is standardized by the __________ of the sample mean.
The pooled proportion is calculated by combining the numbers of successes from the __________ samples.
The Central Limit Theorem applies when observations are independent and the population has a finite __________.
Short-answer questions
Define a large sample.
What is a large-sample test?
State the Central Limit Theorem.
Explain the importance of the Central Limit Theorem in large-sample testing.
What are the assumptions of the Central Limit Theorem stated in this chapter?
Define the Z-test.
State the four large-sample tests discussed in the chapter.
Explain the general decision rule for a Z-test.
What is the difference between a one-tailed and a two-tailed Z-test?
Explain the test for a single population proportion.
State the null and alternative hypotheses for a single population proportion.
Explain why \(P_0\) is used in the standard error of the single-proportion test.
What is the test for equality of two population proportions?
Explain the pooled proportion used in the test for equality of two proportions.
State the null and alternative hypotheses for equality of two proportions.
Explain the test for a single population mean.
Distinguish between the cases where the population standard deviation is known and unknown.
State the test statistic for a single population mean when \(\sigma\) is known.
State the test statistic for a single population mean when \(\sigma\) is unknown.
Explain the test for equality of two population means.
Distinguish between the equal and unequal population standard deviation cases in the test for two means.
State the test statistic when the two population standard deviations are equal.
State the test statistic when the two population standard deviations are unequal.
What is the pooled estimate of the common standard deviation when the population standard deviations are equal but unknown?
Explain how the critical value is selected for a two-tailed test.
Explain how the critical value is selected for a right-tailed test.
Explain how the critical value is selected for a left-tailed test.
Why is the absolute value of \(Z\) used in a two-tailed test?
Explain the meaning of \(Z\sim N(0,1)\) in large-sample testing.
Why can a Z-test be used even when the population distribution is unknown for a sufficiently large sample?
Numerical and conceptual questions
Answers are given at the end of the chapter.
A sample of 500 apples contains 60 bad apples. Test whether the population proportion of bad apples is 0.1 at the 5% level of significance.
A random sample of 500 plants contains 65 plants affected by yellowing disease. Test whether the population proportion of affected plants differs from 0.1 at \(\alpha=0.05\).
In a survey, 50 out of 80 large-holding farmers and 78 out of 250 small-holding farmers adopted a new paddy variety. Test whether the adoption proportions differ significantly at \(\alpha=0.01\).
Explain why a pooled proportion is used when testing equality of two population proportions.
A population has a hypothesized proportion of 0.20. A sample of 400 observations gives a sample proportion of 0.23. State the hypotheses for testing whether the population proportion differs from 0.20.
A fertilizer packing unit is designed to fill bags with a mean weight of 50 kg. The population standard deviation is known to be 3.2 kg. A sample of 64 bags has a mean weight of 49.2 kg. Test the hypothesis at \(\alpha=0.05\).
A sample of 900 items has mean 3.4 cm and standard deviation 2.61 cm. Test whether the population mean is 3.25 cm at \(\alpha=0.05\).
From a paddy field, a random sample of 36 plants has mean panicle length 18.7 cm and standard deviation 1.25 cm. Test whether the population mean panicle length is 19 cm at \(\alpha=0.05\).
A population has known standard deviation 5. A sample of 100 observations has mean 48. Test whether the population mean is 50 at \(\alpha=0.05\).
A large sample has mean 102, while the hypothesized population mean is 100. The population standard deviation is 10 and the sample size is 400. Calculate the Z statistic and test the hypothesis at \(\alpha=0.05\).
Two large samples have sizes 1000 and 2000, with means 67.5 and 68 inches respectively. The common population standard deviation is 2.5 inches. Test whether the two samples can be regarded as drawn from the same population at the 5% level.
Two paddy varieties have sample sizes 100 each. Variety A has mean yield 52 quintals per hectare and population standard deviation 8 quintals per hectare, while variety B has mean yield 49 quintals per hectare and population standard deviation 6 quintals per hectare. Test whether the two varieties differ significantly in mean yield at \(\alpha=0.05\).
Two random samples have \(n_1=400\), \(n_2=400\), \(\bar{x}_1=250\), \(\bar{x}_2=220\), \(s_1=40\), and \(s_2=55\). Test whether the population means are equal at \(\alpha=0.05\).
A two-tailed test is conducted at \(\alpha=0.05\). State the critical value.
A right-tailed test is conducted at \(\alpha=0.01\). State the critical value.
A left-tailed test is conducted at \(\alpha=0.05\). State the critical value.
A calculated \(Z\) value is 2.30 in a two-tailed test at \(\alpha=0.05\). State the decision.
A calculated \(Z\) value is 1.50 in a two-tailed test at \(\alpha=0.05\). State the decision.
A calculated \(Z\) value is 2.00 in a right-tailed test at \(\alpha=0.05\). State the decision.
A calculated \(Z\) value is \(-2.00\) in a left-tailed test at \(\alpha=0.05\). State the decision.
Important formulae
Central Limit Theorem:
\[ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\xrightarrow{d}N(0,1) \]
Single population proportion:
\[ Z=\frac{p-P_0}{\sqrt{P_0Q_0/n}} \]
Complementary proportion:
\[ Q_0=1-P_0 \]
Pooled proportion for two proportions:
\[ \widehat{P}=\frac{n_1p_1+n_2p_2}{n_1+n_2} \]
Pooled complementary proportion:
\[ \widehat{Q}=1-\widehat{P} \]
Equality of two proportions:
\[ Z=\frac{p_1-p_2}{\sqrt{\widehat{P}\widehat{Q}\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}} \]
Single population mean when \(\sigma\) is known:
\[ Z=\frac{\bar{x}-\mu_0}{\sigma/\sqrt{n}} \]
Single population mean when \(\sigma\) is unknown:
\[ Z=\frac{\bar{x}-\mu_0}{s/\sqrt{n}} \]
Equality of two means when population standard deviations are equal:
\[ Z=\frac{\bar{x}_1-\bar{x}_2}{\sigma\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}} \]
Pooled estimate of common standard deviation:
\[ \sigma=\sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}} \]
Equality of two means when population standard deviations are unequal:
\[ Z=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} \]
Critical values of the standard normal distribution
One-tailed test
| \(\alpha\) (one tail) | \(Z\) value |
|---|---|
| 0.10 | 1.282 |
| 0.05 | 1.645 |
| 0.025 | 1.960 |
| 0.01 | 2.326 |
| 0.005 | 2.576 |
Two-tailed test
| \(\alpha\) (two tail) | \(Z\) value |
|---|---|
| 0.20 | 1.282 |
| 0.10 | 1.645 |
| 0.05 | 1.960 |
| 0.02 | 2.326 |
| 0.01 | 2.576 |
Quick revision
- Large sample → generally \(n\geq30\).
- Large-sample tests are based on the Central Limit Theorem.
- Central Limit Theorem → sampling distribution of \(\bar{X}\) approaches a normal distribution as \(n\) increases.
- The observations should be independent.
- The population should have finite variance.
- Large-sample Z-test → based on \(N(0,1)\).
- Four tests in this chapter → single proportion, equality of two proportions, single mean, and equality of two means.
- Single proportion → compare \(P\) with a specified value \(P_0\).
- For a single proportion, \(H_0:P=P_0\).
- For a single proportion, the alternative may be \(P<P_0\), \(P>P_0\), or \(P\neq P_0\).
- Single-proportion standard error uses \(P_0\) and \(Q_0\) under the null hypothesis.
- Two proportions → compare \(P_1\) and \(P_2\).
- For two proportions, \(H_0:P_1=P_2\).
- The two-proportion test uses the pooled proportion \(\widehat{P}\).
- Single mean → compare \(\mu\) with \(\mu_0\).
- When \(\sigma\) is known, use \(\sigma/\sqrt{n}\).
- When \(\sigma\) is unknown, use \(s/\sqrt{n}\).
- Equality of two means → compare \(\mu_1\) and \(\mu_2\).
- Equal population standard deviations → use the common \(\sigma\).
- Equal but unknown standard deviations → use the pooled estimate.
- Unequal population standard deviations → use separate variance terms.
- Two-tailed test → reject if \(|Z|>Z_{\alpha/2}\).
- Right-tailed test → reject if \(Z>Z_\alpha\).
- Left-tailed test → reject if \(Z<-Z_\alpha\).
- At \(\alpha=0.05\), two-tailed critical value → \(1.960\).
- At \(\alpha=0.01\), two-tailed critical value → \(2.576\).
- At \(\alpha=0.05\), right-tailed critical value → \(1.645\).
- At \(\alpha=0.05\), left-tailed critical value → \(-1.645\).
- The null hypothesis is always stated in terms of the population parameter.
- The test statistic is calculated from sample statistics.
- If the calculated statistic falls in the critical region, reject \(H_0\).
- Otherwise, do not reject \(H_0\).
- For a two-tailed test, \(\alpha\) is divided equally between the two tails.
- The sign of \(Z\) is important in one-tailed tests; \(|Z|\) is used for the two-tailed decision.
- A large absolute value of \(Z\) indicates greater departure from the null hypothesis.
Answers to fill in the blanks
1. 30 2. Central 3. Normal 4. 0; 1 5. Z 6. Standard 7. Means 8. \(P_0\) 9. \(1-P_0\) 10. Null 11. \(\widehat{P}\) 12. \(1-\widehat{P}\) 13. \(\mu_0\) 14. \(\sigma\) 15. Sample 16. \(\mu_2\) 17. \(\sigma\) 18. Two 19. 1.960 20. 2.576 21. 1.645 22. \(-1.645\) 23. Two 24. Greater 25. Greater 26. Less 27. \(p\) 28. \(\bar{x}\) 29. \(P\) 30. \(\mu\) 31. \(\sigma\) 32. \(s\) 33. Standard error 34. Two 35. Variance
Solutions to numerical and conceptual questions
Given \(n=500\), \(p=\frac{60}{500}=0.12\), \(P_0=0.1\), \(Q_0=0.9\). Using Equation 14.2, \(Z=\frac{0.12-0.10}{\sqrt{0.1(0.9)/500}}=1.49\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|1.49|<1.960\), we do not reject \(H_0\). There is insufficient evidence that the proportion of bad apples differs from 0.1.
Given \(n=500\), \(p=\frac{65}{500}=0.13\), \(P_0=0.1\), \(Q_0=0.9\). Using Equation 14.2, \(Z=\frac{0.13-0.10}{\sqrt{0.1(0.9)/500}}=2.24\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|2.24|>1.960\), we reject \(H_0\). There is significant evidence that the proportion of affected plants differs from 0.1.
Given \(n_1=80\), \(n_2=250\), \(p_1=\frac{50}{80}=0.625\), \(p_2=\frac{78}{250}=0.312\). The pooled proportion is \(\widehat{P}=\frac{80(0.625)+250(0.312)}{330}=0.3879\) and \(\widehat{Q}=0.6121\). Using Equation 14.3, \(Z=\frac{0.625-0.312}{\sqrt{0.3879(0.6121)\left(\frac{1}{80}+\frac{1}{250}\right)}}=5.00\). At \(\alpha=0.01\), \(Z_{0.005}=2.576\); since \(|5.00|>2.576\), we reject \(H_0\). There is a significant difference between the adoption proportions of the two groups.
Under \(H_0:P_1=P_2\), the two samples are assumed to come from populations having the same proportion. The pooled proportion \(\widehat{P}=\frac{n_1p_1+n_2p_2}{n_1+n_2}\) provides a common estimate of this population proportion under \(H_0\).
The hypothesized population proportion is 0.20, so \(H_0:P=0.20\). Since the question asks whether the proportion differs from 0.20, \(H_1:P\neq0.20\). This is a two-tailed test.
Given \(\bar{x}=49.2\), \(\mu_0=50\), \(\sigma=3.2\), \(n=64\). Using Equation 14.4, \(Z=\frac{49.2-50}{3.2/\sqrt{64}}=\frac{-0.8}{0.4}=-2.00\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|-2.00|>1.960\), we reject \(H_0\). The packing process is not working at the intended mean weight of 50 kg.
Given \(n=900\), \(\bar{x}=3.4\), \(s=2.61\), \(\mu_0=3.25\). Since \(\sigma\) is unknown, using Equation 14.5, \(Z=\frac{3.4-3.25}{2.61/\sqrt{900}}=1.724\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|1.724|<1.960\), we do not reject \(H_0\). There is insufficient evidence that the population mean differs from 3.25 cm.
Given \(n=36\), \(\bar{x}=18.7\), \(s=1.25\), \(\mu_0=19\). Using Equation 14.5, \(Z=\frac{18.7-19}{1.25/\sqrt{36}}=\frac{-0.3}{0.2083}=-1.44\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|-1.44|<1.960\), we do not reject \(H_0\). There is insufficient evidence that the mean panicle length differs from 19 cm.
Given \(\mu_0=50\), \(\sigma=5\), \(n=100\), \(\bar{x}=48\). Using Equation 14.4, \(Z=\frac{48-50}{5/\sqrt{100}}=\frac{-2}{0.5}=-4\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|-4|>1.960\), we reject \(H_0\). There is significant evidence that the population mean differs from 50.
Given \(\bar{x}=102\), \(\mu_0=100\), \(\sigma=10\), \(n=400\). Using Equation 14.4, \(Z=\frac{102-100}{10/\sqrt{400}}=\frac{2}{0.5}=4\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|4|>1.960\), we reject \(H_0\). There is significant evidence that the population mean differs from 100.
Given \(n_1=2000\), \(n_2=1000\), \(\bar{x}_1=68\), \(\bar{x}_2=67.5\), \(\sigma=2.5\). Using Equation 14.6, \(Z=\frac{68-67.5}{2.5\sqrt{\frac{1}{2000}+\frac{1}{1000}}}=5.16\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|5.16|>1.960\), we reject \(H_0\). The two samples cannot be regarded as drawn from populations having the same mean.
Given \(n_1=n_2=100\), \(\bar{x}_1=52\), \(\bar{x}_2=49\), \(\sigma_1=8\), \(\sigma_2=6\). Since the population standard deviations are unequal, using Equation 14.7, \(Z=\frac{52-49}{\sqrt{\frac{8^2}{100}+\frac{6^2}{100}}}=\frac{3}{1}=3.00\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|3.00|>1.960\), we reject \(H_0\). The two paddy varieties differ significantly in mean yield.
Given \(n_1=n_2=400\), \(\bar{x}_1=250\), \(\bar{x}_2=220\), \(s_1=40\), \(s_2=55\). Since the population standard deviations are not given as known, use the pooled estimate \(\sigma=\sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2}}=\sqrt{\frac{400(40^2)+400(55^2)}{800}}=\sqrt{2312.5}\approx48.09\). Using Equation 14.6, \(Z=\frac{250-220}{48.09\sqrt{\frac{1}{400}+\frac{1}{400}}}\approx8.82\). At \(\alpha=0.05\), \(Z_{0.025}=1.960\); since \(|8.82|>1.960\), we reject \(H_0\). There is significant evidence that the two population means are different.
For a two-tailed test at \(\alpha=0.05\), \(\frac{\alpha}{2}=0.025\), so \(Z_{0.025}=1.960\). The critical values are \(-1.960\) and \(1.960\).
For a right-tailed test at \(\alpha=0.01\), \(Z_{0.01}=2.326\); reject \(H_0\) if \(Z>2.326\).
For a left-tailed test at \(\alpha=0.05\), \(Z_{0.05}=1.645\), so the critical value is \(-1.645\); reject \(H_0\) if \(Z<-1.645\).
For a two-tailed test at \(\alpha=0.05\), \(Z_{0.025}=1.960\). Given \(Z=2.30\), since \(|2.30|>1.960\), we reject \(H_0\).
Given \(Z=1.50\). Since \(|1.50|<1.960\), we do not reject \(H_0\).
For a right-tailed test at \(\alpha=0.05\), \(Z_{0.05}=1.645\). Given \(Z=2.00\), since \(2.00>1.645\), we reject \(H_0\).
For a left-tailed test at \(\alpha=0.05\), \(-Z_{0.05}=-1.645\). Given \(Z=-2.00\), since \(-2.00<-1.645\), we reject \(H_0\).
Why 30? The bell curve, the astronomer, and the birth of large-sample theory
The reason large-sample tests work at all traces back to one of the most remarkable results in statistics: the Central Limit Theorem. It says that no matter what the shape of the parent population, the distribution of the sample mean approaches a normal distribution as the sample size grows. This is why the \(Z\)-test, which assumes normality, can be applied to almost any large sample.
The normal curve itself was first derived by Abraham de Moivre in 1733 as an approximation to the binomial distribution, while he was studying games of chance. It was later rediscovered by Carl Friedrich Gauss around 1809, who used it to model the errors astronomers made when measuring the positions of stars and planets. Because the same bell-shaped curve kept appearing whenever many small independent errors were added together, it came to be called the normal distribution, and for a long time the Gaussian distribution. (Gauss 1809)
But where did the famous cut-off of “\(n \geq 30\)” come from? It was never a law of nature. It emerged as a practical rule of thumb in the early twentieth century, around the time William Gosset (writing under the pen name Student) developed the \(t\)-distribution in 1908 while working as a brewer at Guinness in Dublin. Gosset showed that for small samples the normal approximation is unreliable and a separate distribution is needed. As the sample size increases, the \(t\)-distribution steadily approaches the normal curve, and by about 30 observations the two are so close that the difference rarely matters in practice. That convenient meeting point gave us the boundary between “small” and “large” samples that we still use today. (Student 1908)
So the next time you check whether your sample size crosses 30, remember that you are standing on the shoulders of a gambler, an astronomer, and a brewer.
“It is a capital mistake to theorize before one has data.”
- Sir Arthur Conan Doyle