Question

Historically, the average score of PGA golfers for one round is 71.4 with a standard deviation of 3.29. A random sample of 102 golfers is taken. What is the probability that the sample mean is less than 72?

	1) 0.0327
	2) 0.4276
	3) 0.2246
	4) 0.5724
	5) 0.9673

Professors in the Economics Department at Western want to determine how challenging the program was for students. Out of a random sample of 21 students, 16 indicated that the program was either "challenging" or "very challenging". The 95% confidence interval estimating the proprotion of all students in the department who thought the program was challenging is given by which of the following?

Question 9 options:

	1) ( 0.60903 , 0.91478 )
	2) ( -0.57974 , 0.94407 )
	3) ( 0.66896 , 0.85485 )
	4) ( 0.57974 , 0.94407 )
	5) ( 0.05593 , 0.42026 )

Answer 1

Solution :

Given that ,

mean = = 71.4

standard deviation = = 3.29

n = 102

= 71.4

= / n = 3.29/ 102 = 0.3257

P( < 72) = P(( - ) / < (72 - 71.4) / 0.3257)

= P(z < 1.8422)

= 0.9673

probability = 0.9673

option 5) is correct

n = 21

x = 16

Point estimate = sample proportion = = x / n = 16 / 21 = 0.76190

1 - = 1 - 0.76190 = 0.2381

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.76190 * 0.2381) / 21)

= 0.18216

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.76190 - 0.18216 < p < 0.76190 + 0.18216

0.57974 < p < 0.94407

The 95% confidence interval for the population proportion p is : 0.57974 , 0.94407

option 4) is correct