Question 12 (1 point) Suppose the national average dollar amount for an automobile insurance claim is $965.92. You work for an agency in Michigan and you are interested in whether or not the state average is less than the national average. The hypotheses for this scenario are as follows: Null Hypothesis: p. a 965.92, Alternative Hypothesis: p. < 965.92. You take a random sample of claims and calculate a p-value of 0.0164 based on the data, what is the appropriate conclusion? Conclude at the 5% level of significance. 0 1) The true average claim amount is significantly less than $965.92. 0 2) The true average claim amount is significantly higher than $965.92. 0 3) The true average claim amount is higher than or equal to $965.92. 0 4i The true average claim amount is significantly different from $965.92. O 5) We did not find enough evidence to say the true average claim amount is less than $965.92.

Question 11 (1 point) You hear on the local news that for the city of Kalamazoo, the proportion of people who support President Trump is 0.54. However, you think it is greater than 0.54. The hypotheses for this test are Null Hypothesis: p s 0.54, Alternative Hypothesis: p > 0.54. If you randomly sample 20 people and 7 of them support President Trump, what is your test statistic and p-value? O 1) Test Statistic: -1.705, P-Value: 1.912 O 2) Test Statistic: 1.705, P-Value: 0.044 O 3) Test Statistic: -1.705, P-Value: 0.956 O 4) Test Statistic: -1.705, P-Value: 0.044 O 5) Test Statistic: 1.705, P-Value: 0.956

Question 10 (1 point) A student at a university wants to determine if the proportion of students that use iPhones is greater than 0.45. If the student conducts a hypothesis test, what will the null and alternative hypotheses be? ( 1) Ho: p > 0.45 HA: P s 0.45 O 2) Ho: p s 0.45 HA: P > 0.45 3) Ho: p = 0.45 HA: P * 0.45 4) Ho: p 2 0.45 HA: P < 0.45 5) Ho: p < 0.45 HA: p 2 0.45

Question 9 (1 point) The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 8 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 8 golfers play a round of golf on his friend's course and records their scores again. The average difference in the scores (treated as the scores on his course - the scores on his friend's course) is 9.582 and the standard deviation of the differences is 15.9274. Calculate a 90% confidence interval to estimate the average difference in scores between the two courses. 0 1) (7.6874,11.4766) O 2) (-0.8895, 20.0535) 0 3) (1.0867, 20.2507) 0 4) (-1.0867, 20.2507) 0 5) (3.9508,15.2132)c

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