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According to a 2016 report from the Institute for College Access and Success \(66 \%\) of all graduates from public colleges and universities had student loans. A public college surveyed a random sample of 400 graduates and found that \(62 \%\) had student loans. a. Test the hypothesis that the percentage of graduates with student loans from this college is different from the national percentage. Use a significance level of \(0.05\). b. After conducting the hypothesis test, a further question one might ask is what proportion of graduates from this college have student loans? Use the sample data to find a \(95 \%\) confidence interval for the proportion of graduates from the college who have student loans. How does this confidence interval support the hypothesis test conclusion?

Short Answer

Expert verified
The hypothesis test and confidence interval calculation help determine whether the proportion of graduates with student loans from the public college aligns with the national average. The specific results will depend on the exact calculations and z-table values used.

Step by step solution

01

Calculation of Test Statistic

Begin by calculating the test statistic. Use the formula for the test statistic: \(Z = (p_1 - p_2) / SE\) where \(SE = \sqrt{ p *(1 - p) * [(1/n_1) + (1/n_2)] }\). Here, \(p_1\) is the sample proportion (0.62), \(p_2\) is the population proportion (0.66), \(SE\) is the standard error, \(p\) is the pooled proportion calculated as \((X_1 + X_2) / (n_1 + n_2)\), \(X_1\) and \(X_2\) are the number of success in two samples, and \(n_1\) and \(n_2\) are the size of two samples.
02

Hypothesis Test

Next, conduct the hypothesis test. The null hypothesis (\(H_0\)) is that the percentage of graduates with student loans is the same as the national average (\(p = p_0\)) and the alternate hypothesis (\(H_a\)) is that the percentage is not the same (\(p \neq p_0\)). First, the test statistic has to be calculated by replacing the values in the formula from Step 1. Then, determine the critical values of \(z\) at 0.025 level of significance (since it's a two-sample test) using the z-table. Compare the calculated \(z\)-value with the critical \(z\)-value to accept or reject the null hypothesis.
03

Confidence Interval Calculation

The confidence interval supports the findings of the hypothesis test by giving an estimated range of values which is likely to include the proportion of all students with student loans from the university. Use the following formula to calculate the confidence interval: \((p \pm z*SE)\), where \(p = 0.62\) (sample proportion), \(SE\) is calculated from Step 1 and \(z\) corresponds to the 95% confidence level (Look up this value using the z-table). The result will be the confidence interval, indicating that we are 95% certain that the true proportion of students with student loans at the university lies within that range.
04

Interpret the Results

Finally, interpret the results. If the calculated \(z\)-value from Step 2 falls in the acceptance region (i.e. it is less than the critical \(z\)-value), then we fail to reject the null hypothesis. In this case, we can say that there is not enough evidence to suggest that the percentage of graduates with student loans from the university is different from the national average. Note how this interpretation matches with the confidence interval calculated in Step 3. If the national average (0.66 or 66%) lies inside this confidence interval, it further reinforces our failure to reject the null hypothesis.

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Key Concepts

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

Confidence Interval
Understanding confidence intervals is crucial when interpreting statistical analyses. A confidence interval is a range of values that is used to estimate the true value of a population parameter. For instance, in the context of the exercise, we are interested in determining what proportion of graduates from a specific college have student loans. By calculating a 95% confidence interval, we're saying that if we were to take many random samples and compute a confidence interval for each, about 95% of these intervals would contain the true proportion.

The importance of a confidence interval lies in its ability to provide a measure of uncertainty around the sample estimate. It will be wider if there is more variability within the data or if we require a higher level of confidence. In the provided exercise, by constructing a 95% confidence interval for the proportion of graduates with student loans, you gain insight into the precision of the sample estimate and how it might generalize to the larger population of all graduates.
Test Statistic Calculation
The test statistic calculation is a pivotal step in hypothesis testing as it provides a method to measure how much the sample data deviates from the null hypothesis. It quantifies the difference between sample statistics and the hypothesized population parameter under the assumption that the null hypothesis is true.

To calculate the test statistic in a proportion hypothesis test, like the one in our exercise about student loans, you need to subtract the population proportion from the sample proportion, then divide by the standard error of the proportion. This formula yields a z-score, which tells us how many standard errors the observed proportion is from the hypothesized proportion. A higher absolute value of the z-score indicates a greater deviation from the null hypothesis, making it more likely to reject the null hypothesis—if it surpasses the critical value determined from the significance level of the test.
Null Hypothesis
The null hypothesis, symbolized as H0, is a foundation of hypothesis testing and represents a statement of no effect or no difference. It is the default assumption that there is no relationship between two measured phenomena. In the case of our example, the null hypothesis asserts that the proportion of graduates with student loans at the college is the same as the national proportion.

When we perform a hypothesis test, we are using sample data to assess whether there is enough evidence to reject the null hypothesis in favor of an alternative hypothesis, which in our exercise is that the actual proportion of student loans at the college is different from the national figure. Significance levels and p-values are used in conjunction with the null hypothesis to determine the likelihood of observing our sample data if the null hypothesis were true. Rejecting the null hypothesis indicates that our sample provides evidence that supports the alternative hypothesis.
Standard Error
The concept of standard error is fundamental to many statistical tests, including hypothesis testing and confidence interval estimation. It represents the standard deviation of the sampling distribution of a statistic; in simpler terms, it's an estimate of the variation we would expect to see in that statistic from sample to sample if we were to repeat our sampling process many times.

In the context of the provided exercise concerning student loans, the standard error aids in understanding the variability of the sample proportion when estimating the population parameter. It is calculated using both the sample proportion and the size of the sample. A smaller standard error indicates that the sample mean is a more precise estimator of the population mean. Consequently, the standard error is a critical component in the process of both the calculation of the test statistic, which helps us to perform the hypothesis test, and the construction of the confidence interval.

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Most popular questions from this chapter

Is it acceptable practice to look at your research results, note the direction of the difference, and then make the alternative hypothesis one-sided in order to achieve a significant difference? Explain.

The mother of a teenager has heard a claim that \(25 \%\) of teenagers who drive and use a cell phone reported texting while driving. She thinks that this rate is too high and wants to test the hypothesis that fewer than \(25 \%\) of these drivers have texted while driving. Her alternative hypothesis is that the percentage of teenagers who have texted when driving is less than \(25 \%\).$$\begin{aligned} &\mathrm{H}_{0}: p=0.25 \\\&\mathrm{H}_{\mathrm{a}}: p<0.25\end{aligned}$$ She polls 40 randomly selected teenagers, and 5 of them report having texted while driving, a proportion of \(0.125 .\) The p-value is \(0.034\). Explain the meaning of the p-value in the context of this question.

A psychologist is interested in testing whether offering students a financial incentive improves their video-game-playing skills. She collects data and performs a hypothesis test to test whether the probability of getting to the highest level of a video game is greater with a financial incentive than without. Her null hypothesis is that the probability of getting to this level is the same with or without a financial incentive. The alternative is that this probability is greater. She gets a p-value from her hypothesis test of \(0.003 .\) Which of the following is the best interpretation of the p-value? i. The p-value is the probability that financial incentives are not effective in this context. ii. The p-value is the probability of getting exactly the result obtained, assuming that financial incentives are not effective in this context. iii. The p-value is the probability of getting a result as extreme as or more extreme than the one obtained, assuming that financial incentives are not effective in this context. iv. The p-value is the probability of getting exactly the result obtained, assuming that financial incentives are effective in this context. \(\mathrm{v}\). The p-value is the probability of getting a result as extreme as or more extreme than the one obtained, assuming that financial incentives are effective in this context.

A researcher studying extrasensory perception (ESP) tests 300 students. Each student is asked to predict the outcome of a large number of coin flips. For each student, a hypothesis test using a \(5 \%\) significance level is performed. If the \(\mathrm{p}\) -value is less than or equal to \(0.05\), the researcher concludes that the student has ESP. Assuming that none of the 300 students actually have ESP, about how many would you expect the researcher to conclude do have ESP? Explain.

In 2017 the Pew Research Center conducted a survey on family-leave practices and attitudes. Respondents were asked to complete this sentence: "When a family member has a serious health condition, caregiver responsibilities ..." with choices being "mainly on women," "mainly on men," or "on both men and women equally." The percentages for each response are shown in the table below. For these age groups, the responses fell only into the two categories shown in the table. Assume a sample size of 1200 for each age group. $$\begin{array}{|lcc|}\hline \text { Age } & \begin{array}{l}\text { Fall mainly } \\ \text { on women }\end{array} & \begin{array}{l} \text { Fall equally on } \\\\\text { men and women }\end{array} \\\\\hline 30-49 & 60 \% & 40 \% \\\50-64 & 62 \% & 38 \%\end{array}$$ Can we conclude that there is a difference in the proportion of people aged 30 to 49 and aged 50 to 64 who feel the primary caregiver responsibility falls on women? Use a significance level of \(0.05\).

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