Question

The owner of a local phone store wanted to determine how much customers are willing to spend on the purchase of a new phone. In a random sample of 10 phones purchased that day, the sample mean was $473.426 and the standard deviation was $29.1709. Calculate a 95% confidence interval to estimate the average price customers are willing to pay per phone.

Question 1 options:

1)	( 452.558 , 494.294 )

2)	( -452.558 , 494.294 )

3)	( 464.201 , 482.651 )

4)	( 452.873 , 493.979 )

5)	( 471.164 , 475.688 )

Question 2 (1 point)

The owner of a local golf course wanted to determine the average age (in years) of the golfers that played on the course. In a random sample of 27 golfers that visited his course, the sample mean was 47 years old and the standard deviation was 5.11 years. Using this information, the owner calculated the confidence interval of (45.3, 48.7) with a confidence level of 90% for the average age. Which of the following is an appropriate interpretation of this confidence interval?

Question 2 options:

1)	We are 90% confident that the average age of the golfers surveyed is between 45.3 and 48.7 years old.

2)	We are 90% confident that the average age of all golfers that play on the golf course is between 45.3 and 48.7 years old.

3)	We cannot determine the proper interpretation of this interval.

4)	We are certain that 90% of the average ages of all golfers will be between 45.3 and 48.7 years old.

5)	We are 90% confident that the proportion of the ages of all golfers is between 45.3 and 48.7 years old.

Answer 1

Answer:

1.

Given,

Sample n = 10

Mean = 473.426

Standard deviation = 29.1709

Significance level = 0.05

alpha/2 = 0.025

degree of freedom = n - 1

= 10 - 1

= 9

t(alpha/2, df) = t(0.025 , 9) = 2.262

Interval = xbar +/- t*s/sqrt(n)

substitute values

= 473.426 +/- 2.262*29.1709/sqrt(10)

= 473.426 +/- 20.8662

= (452.5598 , 494.2922)

Option A is right answer.

2.

We are 90% confident that the average age of all golfers that play on the golf course is between 45.3 and 48.7 years old.

i.e.,

Option B is right answer.