The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 21 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 21 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 -1.93 strokes and the standard deviation of the differences is 8.674 strokes. The owner uses this information to calculate the 90% confidence paired t-interval of (-5.195, 1.335). Which statement is the correct interpretation of this interval?
Question 11 options:
|
|||
|
|||
|
|||
|
|||
|
Ans :
We are 90% confident that the average difference in scores between the two courses for all golfers is between -5.195 and 1.335.
The owner of a golf course wants to determine if his golf course is more difficult...
The owner of a golf course wants to determine if his golf course is more difficult than the one his friend owns. He has 27 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 27 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...
1.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...
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...
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 10 golfers play a round of 18 holes on his golf course and records their scores. Later that week, he has the same 10 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...
A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 18 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after - before) was -8.603 mg/dL with a standard deviation of 5.449 mg/dL. When creating a 90% confidence interval for the true average difference in cholesterol levels by the drug, what is the margin of error? 1) 1.7087 2) 2.2271 3) 2.2342 4) 1.2843 5) 2.1127...
Question 7 (1 point) The owner of a local golf course wants to determine the average age of the golfers that play on the course in relation to the average age in the area. According to the most recent census, the town has an average age of 55.89. In a random sample of 30 golfers that visited his course, the sample mean was 45.84 and the standard deviation was 6.514. Using this information, the owner calculated the confidence interval of...
The owner of a local golf course wants to estimate the difference between the average ages of males and females that play on the golf course. He samples a group of men and women and then uses the sample statistics to calculate a 99% confidence interval of (6.28, 21.77). This interval estimates the difference of (the average age of men - the average age of women). What can we conclude from this interval? Question 8 options: 1) We are 99%...
(10 pts) Use the following information for all parts: On a particular golf course, a sample of 40 golfers have a mean golf score of 79. Suppose the population standard deviation for this course is 3.6605. (a) Using the formula a 90% confidence interval as presented in lecture, fill in the blanks with the appropriate values for this problem for calculating the confidence interval below. To enter where x is any number, type sqrt(x). For example, should be typed as...
Question 1 (1 point) A customer wants to estimate the average delivery time of a pizza from the local pizza parlor. Over the course of a few months, the customer orders 29 pizzas and records the delivery times. The average delivery time is 23.69 with a standard deviation of 7.212. If the customer estimates the time using a 95% confidence interval, what is the margin of error? Question 1 options: 1) 0.7137 2) 2.2782 3) 2.7433 4) 1.3392 5) 2.739...
The owner of a local golf course wants to examine the difference between the average ages of males and females that play on the golf course. Specifically, he wants to test if the average age of males is greater than the average age of females. If the owner conducts a hypothesis test for two independent samples and calculates a p-value of 0.0095, what is the appropriate conclusion? Label males as group 1 and females as group 2. Question 15 options:...