/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 According to a June 2009 report ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to a June 2009 report (http://www.alertnet.org/thenews/newsdesk/L31011082.htm), \(68 \%\) of people with "green" jobs in North America felt that they had job security, whereas \(60 \%\) of people with green jobs in the United Kingdom felt that they had job security. Suppose that these results were based on samples of 305 people with green jobs from North America and 280 people with green jobs from the United Kingdom. a. Construct a \(96 \%\) confidence interval for the difference between the two population proportions. b. Using the \(2 \%\) significance level, can you conclude that the proportion of all people with green jobs in North America who feel that they have job security is higher than the corresponding proportion for the United Kingdom? Use the critical-value approach. c. Repeat part b using the \(p\) -value approach.

Short Answer

Expert verified
The answers highly depend on the computation based on the given sample proportions and sample sizes. In general, after constructing the confidence interval in step 2, if it does not include zero, we have evidence that the proportions are different. In steps 3 and 4, if the null hypothesis is rejected, we conclude that the proportion of people with 'green' jobs feeling job security in North America is higher than in the UK at a 2% significance level.

Step by step solution

01

Calculate the Sample Proportions

First, we need to get the sample proportions from each population. We compute these by using the given percentages and the sample sizes. \nFor North America: \( p_1 = \frac{68}{100} \times 305 = 207.4 \)\nFor the United Kingdom: \( p_2 = \frac{60}{100} \times 280 = 168 \)
02

Compute the Confidence Interval

The formula for a confidence interval for the difference in population proportions is \[\( (\hat{p}_1 - \hat{p}_2) \pm Z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \) \]\nHere, \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions, \(n_1\) and \(n_2\) are the sample sizes, and \(Z_{\alpha/2}\) is the critical value for the desired level of confidence.\nWe substitute the given values and calculate the confidence interval.
03

Hypothesis Testing using Critical Value Approach

We'll set up our hypotheses as follows:\nNull Hypothesis (H0): \(p_1 - p_2 = 0\) (The proportions are the same)\nAlternative Hypothesis (H1): \(p_1 - p_2 > 0\) (The proportion in North America is greater than in the UK)\nWe calculate the test statistic using the given formula and compare it with the critical value (Z_{\alpha/2}) for a 2% significance level. If our test statistic is greater than the critical value, we reject the null hypothesis.
04

Hypothesis Testing using P-value Approach

In this step, we perform the same hypothesis test, but using the p-value approach. First, we calculate the p-value from our test statistic. After that, we compare the p-value with our significance level (0.02). If the p-value is less than the significance level, we reject the null hypothesis and conclude that the proportion in North America is greater.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Confidence Interval
When we talk about confidence intervals, we are discussing a statistical method used to estimate the true difference between two population proportions. This approach gives us a range, derived from our sample data, within which we are reasonably certain the true difference lies. It offers insight into the variability inherent in any sampling process.

To calculate a confidence interval for the difference in population proportions, we first determine the sample proportions. For example, if \(\hat{p}_1\) represents the proportion of North American workers feeling job security, and \(\hat{p}_2\) represents the same for the UK, our interval will be centered around the difference, \(\hat{p}_1 - \hat{p}_2\).

The formula used is:
  • ***\( (\hat{p}_1 - \hat{p}_2) \pm Z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \)***
Here, \(Z_{\frac{\alpha}{2}}\) is a critical value from the standard normal distribution that corresponds to our desired confidence level, in this case, 96%. This interval provides a range wherein the difference between the population proportions is likely to lie.
Population Proportions
In statistical terms, population proportions refer to the fraction of individuals in a particular group who share a specific characteristic. These proportions help in comparing various populations. For instance, we explored the difference between the proportion of workers with ‘green’ jobs who feel secure in North America and the UK.

To find these proportions, we use sample data. In a sample of North American workers feeling secure, \(\hat{p}_1\), is calculated as:
  • Number of people with job security / Total sample size
Similarly, \(\hat{p}_2\) is computed for UK workers. Calculating these sample proportions allows us to make inferences about the larger population.

Understanding population proportions is fundamental to hypothesis testing, where they're often compared across different groups to discern differences.
Significance Level
In hypothesis testing, the significance level (\(\alpha\)) determines the threshold for rejecting a null hypothesis. It reflects our tolerance for making a Type I error, which occurs when we wrongly reject a true null hypothesis.

A common choice is 5% (\(\alpha = 0.05\)), but in our exercise, we used a more stringent significance level of 2%. This means we want only a 2% risk of incorrectly concluding that the proportion of job security in North America is higher than in the UK when it actually isn't.

Choosing a lower significance level like 2% makes it harder to reject the null hypothesis, thereby increasing our confidence in the results if the null is indeed rejected.
Critical Value Approach
The critical value approach involves comparing our test statistic to a critical value to determine if we should reject our null hypothesis. This method is grounded in the distribution of our test statistic under the assumption that the null hypothesis is true.

In our scenario, we set the following hypotheses:
  • Null Hypothesis (\(H_0\)): \(p_1 - p_2 = 0\) (The proportions are equal)
  • Alternative Hypothesis (\(H_1\)): \(p_1 - p_2 > 0\) (North America's proportion is greater)
We then calculate a test statistic based on our sample data and compare it to the critical value from the z-distribution that aligns with our chosen significance level.

If our test statistic exceeds this critical value, we reject the null hypothesis. This implies sufficient evidence to support the claim that more North Americans in green jobs feel secure compared to their UK counterparts.
P-value Approach
The p-value approach offers a different route to hypothesis testing by quantifying the evidence against the null hypothesis. A p-value is the probability of observing data as extreme as, or more so than, our sample data, under the null hypothesis.

In our exercise, after calculating the test statistic, we determine the corresponding p-value. We then compare this p-value to our significance level of 2% (0.02).
  • If the p-value ≤ significance level, we reject the null hypothesis, suggesting a significant difference in job security proportions between North America and the UK.
  • If the p-value > significance level, we cannot reject the null hypothesis.
The p-value approach provides a specific level of evidence against the null hypothesis, making it intuitive and informative for decision-making in hypothesis testing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Does the use of cellular telephones increase the risk of brain tumors? Suppose that a manufacturer of cell phones hires you to answer this question because of concern about public liability suits. How would you conduct an experiment to address this question? Be specific. Explain how you would observe, how many observations you would take, and how you would analyze the data once you collect them. What are your null and alternative hypotheses? Would you want to use a higher or a lower significance level for the test? Explain.

A mail-order company has two warehouses, one on the West Coast and the second on the East Coast. The company's policy is to mail all orders placed with it within 72 hours. The company's quality control department checks quite often whether or not this policy is maintained at the two warehouses. A recently taken sample of 400 orders placed with the warehouse on the West Coast showed that 364 of them were mailed within 72 hours. Another sample of 300 orders placed with the warehouse on the East Coast showed that 279 of them were mailed within 72 hours.

Construct a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) for the following. $$ n_{1}=100 \quad \hat{p}_{1}=.81 \quad n_{2}=150 \quad \hat{p}_{2}=.77 $$

The following information was obtained from two independent samples selected from two populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=55 & \bar{x}_{1}=90.40 & s_{1}=11.60 \\ n_{2}=50 & \bar{x}_{2}=86.30 & s_{2}=10.25 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

As was mentioned in Exercise \(9.38\), The Bath Heritage Days, which take place in Bath, Maine, switched to a Whoopie Pie eating contest in 2009 . Suppose the contest involves eating nine Whoopie Pies, each weighing \(1 / 3\) pound. The following data represent the times (in seconds) taken by each of the 13 contestants (all of whom finished all nine Whoopie Pies) to eat the first Whoopie Pie and the last (ninth) Whoopie Pie. $$ \begin{array}{l|rrrrrrrrrrrr} \hline \text { Contestant } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} & \mathbf{1 0} & \mathbf{1 1} & \mathbf{1 2} & \mathbf{1 3} \\ \hline \text { First pie } & 49 & 59 & 66 & 49 & 63 & 70 & 77 & 59 & 64 & 69 & 60 & 58 & 71 \\ \hline \text { Last pie } & 49 & 74 & 92 & 93 & 91 & 73 & 103 & 59 & 85 & 94 & 84 & 87 & 111 \\ \hline \end{array} $$ a. Make a \(95 \%\) confidence interval for the mean of the population paired differences, where a paired difference is equal to the time taken to eat the ninth pie (which is the last pie) minus the time taken to eat the first pie. b. Using the \(10 \%\) significance level, can you conclude that the average time taken to eat the ninth pie (which is the last pie) is at least 15 seconds more than the average time taken to eat the first pie.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.