Question

In a group of 15 young professionals studied, it was found that they purchased on average $186.498 per month eating meals prepared outside the home, with a standard deviation of $30.392. What is the 99% confidence interval of the true amount of money young professionals spend monthly on meals prepared outside the home?

	1) ( 163.138 , 209.858 )
	2) ( 183.521 , 189.475 )
	3) ( 163.372 , 209.624 )
	4) ( -163.138 , 209.858 )
	5) ( 178.651 , 194.345 )

You read an article in Golf Digest that someone with your average drive distance will have an average score of 79.53. You keep track of your scores for the next 13 rounds and see that your average score is 79.2 with a standard deviation of 1.295 strokes. You create a 95% confidence interval for your average score and it is (78.42, 79.98). Given this information, what is the best conclusion?

	1) The percentage of rounds in which you score more than 79.53 is 95%.
	2) We cannot determine the proper interpretation based on the information given.
	3) The average score does not significantly differ from 79.53.
	4) We are 95% confident that your average score is greater than 79.53.
	5) We are 95% confident that your average score is less than 79.53.

It is believed that using a solid state drive (SSD) in a computer results in faster boot times when compared to a computer with a traditional hard disk (HDD). You sample a group of computers and use the sample statistics to calculate a 95% confidence interval of (1.64, 10.4). This interval estimates the difference of (average boot time (HDD) - average boot time (SSD)). What can we conclude from this interval?

	1) We are 95% confident that the average boot time of all computers with an HDD is greater than the average of all computers with an SSD.
	2) We are 95% confident that the average boot time of all computers with an SSD is greater than the average of all computers with an HDD.
	3) There is no significant difference between the average boot time for a computer with an SSD drive and one with an HDD drive at 95% confidence.
	4) We do not have enough information to make a conclusion.
	5) We are 95% confident that the difference between the two sample means falls within the interval.

A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 8 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after - before) was 15.539 mg/dL with a standard deviation of 19.2612 mg/dL. Create a 90% confidence interval for the true average difference in cholesterol levels by the drug.

	1) (-2.6372, 28.4408)
	2) (2.6372, 28.4408)
	3) (13.6444, 17.4336)
	4) (2.8757, 28.2023)
	5) (8.7291, 22.3489)

As of 2012, the proportion of students who use a MacBook as their primary computer is 0.31. You believe that at your university the proportion is actually less than 0.31. If you conduct a hypothesis test, what will the null and alternative hypotheses be?

	1) HO: p > 0.31 HA: p ≤ 0.31
	2) HO: p ≥ 0.31 HA: p < 0.31
	3) HO: p ≤ 0.31 HA: p > 0.31
	4) HO: p = 0.31 HA: p ≠ 0.31
	5) HO: p < 0.31 HA: p ≥ 0.31

Answer 1

n= 15 x=0.01 trol = 2.9768 = 186.498 s = 30.392 99%. C.I for a ł I twiga/2 s in 186.498 + 2.976844 30.392 VIS (163.138 , 209.858) option A