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Chapter 11: Problem 9

In Problems 9鈥�12, conduct each test at the a = 0.05 level of significance by determining (a) the null and alternative hypotheses, (b) the test statistic, (c) the critical value, and (d) the P-value. Assume that the samples were obtained independently using simple random sampling. Test whether \(p_{1}>p_{2}\). Sample data: \(x_{1}=368, n_{1}=541\), \(x_{2}=351, n_{2}=593\)

Short Answer

Expert verified
Reject the null hypothesis because the P-value (0.0066) is less than 0.05, suggesting \( p_1 > p_2 \).

Step by step solution

01

Identify Null and Alternative Hypotheses

Define the null hypothesis and alternative hypothesis. Let鈥檚 denote the proportions as follows: - Null hypothesis, \( H_0 \): \( p_1 \le p_2 \)- Alternative hypothesis, \( H_a \): \( p_1 > p_2 \)
02

Calculate the sample proportions

The sample proportions can be calculated as follows: \( \hat{p}_1 = \frac{x_1}{n_1} \) and \( \hat{p}_2 = \frac{x_2}{n_2} \). - \( \hat{p}_1 = \frac{368}{541} \approx 0.680 \)- \( \hat{p}_2 = \frac{351}{593} \approx 0.592 \)
03

Determine the test statistic

The formula for the test statistic when comparing two proportions is given by \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{ \hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right) }} \]where \( \hat{p} \) is the pooled sample proportion: \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \).First, calculate the pooled sample proportion: - \( \hat{p} = \frac{368 + 351}{541 + 593} = \frac{719}{1134} \approx 0.634 \)Next, plug in the values to get the test statistic: - \( z = \frac{(0.680 - 0.592)}{\sqrt{0.634 \times (1 - 0.634) \times \left(\frac{1}{541} + \frac{1}{593}\right)}} \approx 2.48 \)
04

Determine the critical value

For a significance level of \( \alpha = 0.05 \) for a one-tailed test (right-tail), refer to the z-table to find that the critical value \( z_{\alpha} = 1.645 \).
05

Calculate the P-value

Using the z-table, find the P-value corresponding to the test statistic \( z = 2.48 \). The P-value is the probability that a standard normal variable is greater than 2.48. From the z-table, P-value \( \approx 0.0066 \).
06

Compare P-value to significance level

Compare the P-value to the significance level \( \alpha = 0.05 \). Since \(0.0066 < 0.05\), we reject the null hypothesis.
07

Conclusion

There is sufficient evidence to conclude that \( p_1 > p_2 \) at the 0.05 level of significance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

null hypothesis
The null hypothesis, often denoted as \(H_0\), is a statement that there is no effect or no difference. In hypothesis testing, it's what we assume to be true until we have evidence to suggest otherwise. For the given exercise, the null hypothesis is \(p_1 \leq p_2\). This means that we initially assume the proportion of success in the first sample (\(p_1\)) is less than or equal to the proportion of success in the second sample (\(p_2\)). We need significant evidence from the data to reject this null hypothesis.
alternative hypothesis
Opposite to the null hypothesis, the alternative hypothesis, denoted as \(H_a \), proposes what we are trying to demonstrate. For our problem, the alternative hypothesis is \(p_1 > p_2\). This suggests that the proportion of success in the first sample is greater than in the second sample. In hypothesis testing, proving the alternative hypothesis means we have strong evidence that there is a significant effect or difference.
test statistic
A test statistic is a standardized value used to decide whether to reject the null hypothesis. It's calculated from your sample data. In our example, the test statistic (z) is determined using a specific formula that incorporates the sample proportions and the pooled proportion. The formula for comparing two proportions is: \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{ \hat{p}(1 - \hat{p}) \left(\frac{1}{n_1} + \frac{1}{n_2}\right) }} \] Here, we first calculate the pooled proportion (\(\hat{p}\)), then substitute the values into the formula to get the test statistic. In our case, it鈥檚 about 2.48.
critical value
The critical value is a threshold that the test statistic must exceed to reject the null hypothesis. It's determined by the significance level (alpha) and the type of test (one-tailed or two-tailed). For a significance level of 0.05 and a one-tailed test, the critical value for z is 1.645. If our test statistic (z) is greater than this critical value, we reject the null hypothesis. Since our calculated test statistic is 2.48, which is greater than 1.645, we have enough evidence against the null hypothesis.
P-value
The P-value represents the probability of obtaining a test statistic at least as extreme as the one computed, assuming the null hypothesis is true. In simpler terms, it helps us measure the strength of the evidence against the null hypothesis. For our exercise, the P-value corresponding to the test statistic (z = 2.48) is approximately 0.0066. We compare this P-value with our significance level (0.05). Since 0.0066 is much smaller than 0.05, we reject the null hypothesis. This small P-value indicates strong evidence that \(p_1 > p_2\).

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Most popular questions from this chapter

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