Question

In: Statistics and Probability

The distribution of GPA scores is known to be left-skewed. At a large university, the administration...

The distribution of GPA scores is known to be left-skewed. At a large university, the administration is interested in learning about the average GPA score of the undergraduate students. A simple random sample of 75 undergraduates results in an average GPA score of 2.97. Assume that the distribution of GPA scores of the undergraduates at this university is also left-skewed with a standard deviation of 0.62.

(a) Which of the following statements is true?

A. The sampling distribution is normal.

B. The population distribution is normal.

C. Both the sampling distribution and the population distribution are normal.

D. Neither the sampling distribution nor the population distribution is normal.

E. Unable to determine with the information provided.

(b) What is the 97% confidence interval for the mean GPA of the undergraduates?

A. (1.625, 4.315)

B. (2.835,3.105)

C. (1.73, 4.70)

D. (2.815, 3.125)

E. (1.804, 4.136)

F. (2.827, 3.113)

G. (2.830, 3.110)

H. None of Above

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 2.97

Population standard deviation = = 0.62

Sample size = n = 75

a) A. The sampling distribution is normal.

b) At 97% confidence level

= 1 - 97%

= 1 - 0.97 =0.03

/2 = 0.015

Z/2 = Z0.015 = 2.17

Margin of error = E = Z/2 * ( / $\sqrt$ n)

= 2.17 * ( 0.62 / $\sqrt$ 75 )

= 0.155

At 97% confidence interval estimate of the population mean is,

± E

2.97 ± 0.155

( 2.815, 3.125 )

correct option is =D

orchestra answered 11 months ago

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