Assume a researcher performs a hypothesis test and uses the below screenshot from StatKey to calculate the p-value: Look at all of the provided information on the graph when answering this question.  The alternative hypothesis of Ha: [answer1] is the only possibility because the p-value was obtained from the theoretical [answer2] distribution and the test is [answer3].

Consider the three histograms shown below. Each are based on n = 900 observations.   Match the histograms to the corresponding boxplots.

A researcher wants to answer the question: Do used Subaru Foresters cost more in Washington State than they do in Tennessee, on the average?  The researcher randomly chooses 20 used Foresters for sale in Washington State and randomly chooses 20 used Foresters for sale in Tennessee. The asking price (in thousands of dollars) and state are recorded for each car. The researcher used the data to create the randomization distribution below:   Decide whether each of the statements below is true or false when considering all provided information.

Data was obtained from a randomly selected sample of n = 22 smart watch users.  These smart watch users indicated they stood for an average of 14 minutes per day with a standard deviation of 4 minutes per day. The calculated standard error of the mean is 0.85 minutes per day. Assume that all conditions are met to use the theoretical t distribution for inference. Which is the correct 99% confidence interval for the average minutes per day stood by Smart-watch users?  

Research Question:  Does the data suggest that less than 60% of Americans have at least $1000 in the bank for an emergency. The sample statistic of interest is p-hat: the proportion of Americans who do have at least $1000 in the bank for an emergency. Company A surveys 240 Americans with a sample proportion of 0.58. Company B surveys 410 Americans with a sample proportion of 0.58. Company C surveys 80 Americans with a sample proportion of 0.58. Which Company has the most evidence against the null hypothesis, i.e. in favor of the research question?

Suppose researchers surveyed 52 expecting mothers at a “Future Mothers” class on their attitudes on the use of formula vs. breastfeeding. The survey results found that 20% of expecting mothers favored the use of formula over breastfeeding.  Using 5,000 Bootstrap samples, we calculate a 95% confidence interval to be 18.6% to 21.4% . Indicate what would happen to the width of the confidence interval if the changes below were made.  For each option, assume all other quantities are held constant.

Research Question: Since 2013 about 3 in 100 (or 3% of) births have included twins. Is there evidence that more than 3% of births in Pennsylvania are twin births? The parameter of interest is a [answer1]. To answer this question, we should [answer2].  

A  study considered the following two variables: 1. Attended office hours: did attend  or did not attend   2. Passed exam 1:  yes    no     yes no total did attend 5  2  7 did not attend 80  13  93 total 85 15  100 Match the statistic to its corresponding calculation

Method 2:  Confidence Intervals calculated using the Percentile Method is not be covered on this exam

1.5 pts Edit this Question Delete this Question Penn State University Health Services (UHS) believes that students who received a flu shot were less likely to have the flu this year when compared to students who did not receive a flu shot. To test this claim, they surveyed a random sample of 200 PSU students. The students were separated by whether or not they had the flu. The sample difference in proportions of students who had the flu, when using the flu shot group first in the calculation, is -0.39. The correct null and alternative hypotheses for this situation would be: Ho :[answer1][answer2][answer3] Ha :[answer4][answer5][answer6] Assuming the relevant p-value for this test is 0.350, complete the conclusion below: We [answer7] the null hypothesis.  We [answer8] convincing evidence that there is a smaller proportion of PSU students who caught the flu this year when comparing students who received a flu shot to those who did not.