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Chapter 9: Problem 44

Independent random samples of 280 and 350 observations were selected from binomial populations 1 and 2 , respectively. Sample 1 had 132 successes, and sample 2 had 178 successes. Do the data present sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population \(2 ?\) Use one of the two methods of testing presented in this section, and explain your conclusions.

Short Answer

Expert verified
Answer: No, there is not enough evidence to conclude that the proportion of successes in population 1 is smaller than the proportion of successes in population 2.

Step by step solution

01

Define Hypotheses and Level of Significance

We will perform a hypothesis test to check if the proportion of successes in population 1 is smaller than that of population 2. The null and alternative hypotheses are as follows: - Null Hypothesis (H0): \(p_1 \geq p_2\) - Alternative Hypothesis (H1): \(p_1 < p_2\) Let's use a 0.05 level of significance for the test.
02

Calculate Sample Proportions

For sample 1, there were 132 successes out of 280 observations, so the sample proportion is: $$\hat{p_1} = \frac{132}{280} = 0.4714$$ For sample 2, there were 178 successes out of 350 observations, so the sample proportion is: $$\hat{p_2} = \frac{178}{350} = 0.5086$$
03

Calculate Pooled Proportion

Since we are assuming that the two proportions are equal under the null hypothesis, we need to calculate the pooled proportion, which is: $$\hat{p} = \frac{132 + 178}{280 + 350} = \frac{310}{630} = 0.4921$$
04

Calculate Standard Error

The standard error for the difference in proportions is given by: $$SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n_1} + \frac{\hat{p}(1-\hat{p})}{n_2}} = \sqrt{\frac{0.4921(1-0.4921)}{280} + \frac{0.4921(1-0.4921)}{350}} = 0.0645$$
05

Calculate Test Statistic

The test statistic for the difference in proportions is given by: $$Z = \frac{(\hat{p_1} - \hat{p_2}) - (p_1 - p_2)}{SE} = \frac{(0.4714 - 0.5086) - 0}{0.0645} = -0.5757$$
06

Calculate P-Value

Now, we need to calculate the p-value, which is the probability of observing a test statistic as extreme or more extreme than the one we calculated, assuming the null hypothesis is true. Since this is a left-tailed test, the p-value is given by: $$P(Z < -0.5757)$$ Looking up this value in a standard normal distribution table or using software, we find that the p-value is approximately 0.2825.
07

Make a Decision

Since the p-value (0.2825) is greater than our level of significance (0.05), we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the proportion of successes in population 1 is smaller than the proportion of successes in population 2.

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Independent random samples of \(n_{1}=140\) and \(n_{2}=140\) observations were randomly selected from binomial populations 1 and \(2,\) respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to which parameter, \(p_{1}\) or \(p_{2}\), is the larger, but you want to detect only a difference between the two parameters if one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, \(\left(\hat{p}_{1}-\hat{p}_{2}\right) .\) Make sure to use the pooled estimate for the common value of \(p\). c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population proportions are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population proportions at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population proportions?

A Cure for Insomnia An experimenter has prepared a drug-dose level that he claims will induce sleep for at least \(80 \%\) of people suffering from insomnia. After examining the dosage we feel that his claim regarding the effectiveness of his dosage are inflated. In an attempt to disprove his claim, we administer his prescribed dosage to 50 insomniacs and observe that 37 of them have had sleep induced by the drug dose. Is there enough evidence to refute his claim at the \(5 \%\) level of significance?

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Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\) If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.
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