Question

The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 11 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 11 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 7.381 and the standard deviation of the differences is 18.7296. Calculate a 95% confidence interval to estimate the average difference in scores between the two courses.

Question 9 options:

	1) (-5.2017, 19.9637)
	2) (-5.0484, 19.8104)
	3) (5.1529, 9.6091)
	4) (5.2017, 19.9637)
	5) (1.7338, 13.0282)

A suggestion is made that the proportion of people who have food allergies and/or sensitivities is 0.51. You believe that the proportion is actually different from 0.51. If you conduct a hypothesis test, what will the null and alternative hypothesis be?

Question 10 options:

	1) HO: p ≤ 0.51 HA: p > 0.51
	2) HO: p ≥ 0.51 HA: p < 0.51
	3) HO: p ≠ 0.51 HA: p = 0.51
	4) HO: p > 0.51 HA: p ≤ 0.51
	5) HO: p = 0.51 HA: p ≠ 0.51

As of 2012, the proportion of students who use a MacBook as their primary computer is 0.3. You believe that at your university the proportion is actually greater than 0.3. The hypotheses for this test are Null Hypothesis: p ≤ 0.3, Alternative Hypothesis: p > 0.3. If you randomly select 20 students in a sample and 8 of them use a MacBook as their primary computer, what is your test statistic and p-value?

Question 11 options:

	1) Test Statistic: 0.976, P-Value: 0.835
	2) Test Statistic: -0.976, P-Value: 0.165
	3) Test Statistic: -0.976, P-Value: 0.835
	4) Test Statistic: 0.976, P-Value: 0.165
	5) Test Statistic: 0.976, P-Value: 0.33

Answer 1

Page No. Date Given n=11, did=7.381 Sdr - 18.7296. 95% confidance interveel to estimate the average difference in scores beturcen tuo Courses. ad & talagnrix Sols 2 7.381 & to.025 10x 18.7296 7.381 t 2:22 818 5.6472 7.381 + 12.5885 - 5.2015 ud 19.9635 (-5.2017 , 19.9637) # Ho: P-0.51- vis Hiu PI 0.51 # R P-0.3 Hoe pao.3 vis Hi P70.3 X -8, nr. 20 Rham 0.4 Test statistic 2 . JP (-p) 0.4-0.3 0.34 0.7 20 - 0.976. P- Valmer P(Zso.976) 0.1645 ~ 0.165