Question

In a random sample of 100 college students, scores on a test were normally distributed with a mean of 80 points and standard deviation of 12 points. Use this scenario to answer the following questions:

A. In this scenario, what is the point estimate? [2 points]

B. Construct a 95% confidence interval to estimate the mean score in the population of all college students. Remember to show all work. Round your final answer to 3 decimal places. [4 points]

C. Interpret the confidence interval you constructed in part B by completing the phrase "I am 95% confident that..." [2 points]

Answer 1

Solution :

Given that,

A)Point estimate = sample mean = = 80

sample standard deviation = s = 12

sample size = n = 100

Degrees of freedom = df = n - 1 = 100 - 1 = 99

B) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,99 = 1.984

Margin of error = E = t/2,df * (s /n)

= 1.984 * (12 / 100)

Margin of error = E = 2.381

The 95% confidence interval estimate of the population mean is,

  ± E

= 80  ± 2.381

= ( 77.619, 82.381 )

C) "I am 95% confident that the mean score in the population of all college students between 77.619 and 82.381.