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A study in the July 7,2009 , issue of \(U S A\) TODAY stated that the \(401(\mathrm{k})\) participation rate among U.S. employees of Asian heritage is \(76 \%\), whereas the participation rate among U.S. employees of Hispanic heritage is \(66 \%\). Suppose that these results were based on random samples of 100 U.S. employees from each group. a. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. b. Using the \(5 \%\) significance level, can you conclude that the \(401(\mathrm{k})\) participation rates are different for all U.S. employees of Asian heritage and all U.S. employees of Hispanic heritage? Use the critical- value and \(p\) -value approaches. c. Repeat parts a and b for both sample sizes of 200 instead of 100 . Does your conclusion change in part b?

Short Answer

Expert verified
After the calculation, it will be seen whether the confidence interval includes zero difference in proportions. If zero is not included in the interval, this suggests that there is a significant difference in the two populations' proportions. Similarly, if the null hypothesis is rejected in the hypothesis test, this also suggests that the proportions are different. The conclusions with the increased sample size will either affirm or challenge the initial findings.

Step by step solution

01

Calculate Sample Proportions and Differences

First, calculate the sample proportions. For Asian: \(p_1 = 0.76\) and For Hispanic: \(p_2 = 0.66\). The difference in sample proportions is \(p_1 - p_2 = 0.76 - 0.66 = 0.10\).
02

Construct a 95% Confidence Interval for the Difference

Formula for confidence interval of difference in proportions is \[p_1 - p_2 \pm Z \sqrt{\frac{{p_1(1 - p_1)}}{{n_1}} + \frac{{p_2(1 - p_2)}}{{n_2}}}\]. Substituting, \(n_1 = 100, n_2 = 100, Z = 1.96\) (for 95% confidence), and the values of \(p_1\) and \(p_2\) from Step 1 into the formula to determine the confidence interval.
03

Perform Hypothesis Testing

The null hypothesis is that the proportions are equal, and the alternative hypothesis is that the proportions are not equal. Perform a Z test of hypothesis for the difference in proportions using the critical value and p-value approaches. If the calculated Z score is greater than the critical Z score, or the p-value is less than the predetermined significance level (0.05), reject the null hypothesis.
04

Repeat Steps 1-3 for a sample size of 200

Now repeat the same processes, but with the sample sizes \(n_1\) and \(n_2\) increased from 100 to 200. This involves recalculating the confidence interval using the increased sample size in the formula (Step 2) and rerunning the hypothesis test (Step 3).
05

Compare the Results

Compare the conclusions reached from the hypothesis tests and confidence intervals calculated with the sample sizes of 100 and 200. If these conclusions are different, then the sample size has an effect on the results of the analysis.

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Key Concepts

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

Understanding Hypothesis Testing
Hypothesis testing is a powerful statistical tool that helps you make inferences about a population based on sample data. It's like playing detective with data, where you try to determine if your findings are significant or just due to random chance.

When you conduct a hypothesis test, you start with two opposing statements:
  • The null hypothesis (often denoted as \(H_0\)): This states that there is no effect or no difference. In our scenario, it suggests that the \(401(k)\) participation rates for Asian and Hispanic employees are the same.
  • The alternative hypothesis (\(H_1\)): This states that there is an effect or a difference. Here, it implies that the participation rates are different for the two groups.
After setting these hypotheses, you go ahead and check your data. You determine if the evidence strongly supports either stance. If your analysis shows that the probability of the null hypothesis being true is too low, you reject \(H_0\) and lean toward \(H_1\).
Exploring Proportions
Proportions tell us how large one part of the whole group is. In the case of our study, proportions show the fraction of Asian and Hispanic employees participating in \(401(k)\) programs.

Here were the sample proportions:
  • For Asian employees, the proportion \(p_1\) is \(0.76\) or 76%.
  • For Hispanic employees, the proportion \(p_2\) is \(0.66\) or 66%.
Understanding these proportions lets us compare and quantify how many employees from each group are participating in the \(401(k)\) plan. This comparison is crucial in constructing confidence intervals and performing hypothesis tests.
The Importance of Sample Size
Sample size, often denoted with \(n\), can greatly influence the results of your analysis. It represents the number of observations or data points collected from a population.

When you increase the sample size, the estimate becomes more precise. A larger sample size can yield more reliable results because it better approximates the true population parameters. In our context:
  • A sample size of 100 was initially used for both groups. With this size, the calculated difference in proportions seemed significant.
  • When the sample size increased to 200, recalculating the confidence interval and hypothesis tests could lead to different results, underlining the impact of sample size on statistical conclusions.
Thus, choosing an adequate sample size is vital for robust and credible statistical inference.
Decoding Significance Level
The significance level, often represented by \(\alpha\), is a critical part of hypothesis testing. It tells us how strong the evidence must be to reject the null hypothesis.

In most studies:
  • A common significance level is \(0.05\), or 5%. This means there's a 5% risk of concluding there is an effect when there truly isn't one.
This level helps set the critical threshold in determining whether our test results are statistically significant.
  • If the p-value (a measure of the evidence against the null hypothesis) is less than \(\alpha\), you reject \(H_0\). It shows that the differences in proportions are unlikely due to chance alone.
  • If the p-value is greater than \(\alpha\), you do not reject \(H_0\), suggesting there's not enough evidence to support that the participation rates differ significantly between the groups.
Selecting the right significance level is key to balancing the risk of incorrect conclusions.

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Most popular questions from this chapter

A factory that emits airborne pollutants is testing two different brands of filters for its smokestacks. The factory has two smokestacks. One brand of filter (Filter I) is placed on one smokestack, and the other brand (Filter II) is placed on the second smokestack. Random samples of air released from the smokestacks are taken at different times throughout the day. Pollutant concentrations are measured from both stacks at the same time. The following data represent the pollutant concentrations (in parts per million) for samples taken at 20 different times after passing through the filters. Assume that the differences in concentration levels at all times are approximately normally distributed. $$ \begin{array}{cccccc} \hline \text { Time } & \text { Filter I } & \text { Filter II } & \text { Time } & \text { Filter I } & \text { Filter II } \\ \hline 1 & 24 & 26 & 11 & 11 & 9 \\ 2 & 31 & 30 & 12 & 8 & 10 \\ 3 & 35 & 33 & 13 & 14 & 17 \\ 4 & 32 & 28 & 14 & 17 & 16 \\ 5 & 25 & 23 & 15 & 19 & 16 \\ 6 & 25 & 28 & 16 & 19 & 18 \\ 7 & 29 & 24 & 17 & 25 & 27 \\ 8 & 30 & 33 & 18 & 20 & 22 \\ 9 & 26 & 22 & 19 & 23 & 27 \\ 10 & 18 & 18 & 20 & 32 & 31 \\ \hline \end{array} $$ a. Make a \(95 \%\) confidence interval for the mean of the population paired differences, where a paired difference is equal to the pollutant concentration passing through Filter I minus the pollutant concentration passing through Filter II. b. Using the \(5 \%\) significance level, can you conclude that the average paired difference for concentration levels is different from zero?

Refer to Exercise \(10.32\). It was mentioned that the average credit limit on 200 credit cards issued during 2008 was \(\$ 4710\) with a standard deviation of \(\$ 485\), and the mean and standard deviation for 200 credit cards issued during the first 4 months of 2009 were \(\$ 4602\) and \(\$ 447\), respectively. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean credit limits for all new credit cards issued in 2008 and during the first 4 months of 2009 . b. Using the \(2.5 \%\) significance level, can you conclude that the average credit limit for all new credit cards issued in 2008 is higher than the corresponding average for the first 4 months of \(2009 ?\) c. Suppose that the standard deviations for the two samples were \(\$ 590\) and \(\$ 257\), respectively. Redo parts a and b. Discuss any changes in the results.

An insurance company wants to know if the average speed at which men drive cars is greater than that of women drivers. The company took a random sample of 27 cars driven by men on a highway and found the mean speed to be 72 miles per hour with a standard deviation of \(2.2\) miles per hour. Another sample of 18 cars driven by women on the same highway gave a mean speed of 68 miles per hour with a standard deviation of \(2.5\) miles per hour. Assume that the speeds at which all men and all women drive cars on this highway are both normally distributed with the same population standard deviation. a. Construct a \(98 \%\) confidence interval for the difference between the mean speeds of cars driven by all men and all women on this highway. b. Test at the \(1 \%\) significance level whether the mean speed of cars driven by all men drivers on this highway is greater than that of cars driven by all women drivers

A company claims that its medicine, Brand \(A\), provides faster relief from pain than another company's medicine, Brand B. A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes. $$ \begin{array}{cccc} \hline & & \text { Mean of } & \text { Standard Deviation } \\ \text { Brand } & \text { Sample Size } & \text { Relief Times } & \text { of Relief Times } \\ \hline \text { A } & 25 & 44 & 11 \\ \text { B } & 22 & 49 & 9 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at the \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B.

As mentioned in Exercise \(10.29\), a company claims that its medicine, Brand A, provides faster relief from pain than another company's medicine, Brand \(\mathrm{B}\). A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes.$$ \begin{array}{cccc} \hline \text { Brand } & \text { Sample Size } & \begin{array}{c} \text { Mean of } \\ \text { Relief Times } \end{array} & \begin{array}{c} \text { Standard Deviation } \\ \text { of Relief Times } \end{array} \\ \hline \text { A } & 25 & 44 & 11 \\ \text { B } & 22 & 49 & 9 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at the \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B. c. Suppose that the sample standard deviations were \(13.3\) and \(7.2\) minutes, respectively. Redo parts a and \(\mathrm{b}\). Discuss any changes in the results.

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