Question

thirty-one randomly selected students scores on the test in MTH 243 were recorded. This sample had a mean of 73 and a standard deviation of 8.4. Construct a 98% confidence interval for the mean score of all students MTH 243 test score. (round all values to three decimal places)

Does this sample satisfy the conditions for constructing a confidence interval?

Select an answer Yes No Why do you keep asking these questions of me?

What critical value is appropriate for this problem?

Select an answer critical t-value critical z-value

What is the value of the critical value? (just enter the number)



What is the margin of error?

E=E=

State the 98% confidence interval.

<μ<<μ<

Answer 1

Yes, this samples are randomly selected i.e samples are iid's

Critical value t(α/2, n-1) = t(0.02 /2, 31- 1 ) = 2.457

Margin of Error = t(α/2, n-1) S/√(n) = 3.707

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.02 /2, 31- 1 ) = 2.457
73 ± t(0.02/2, 31 -1) * 8.4/√(31)
Lower Limit = 73 - t(0.02/2, 31 -1) 8.4/√(31)
Lower Limit = 69.293
Upper Limit = 73 + t(0.02/2, 31 -1) 8.4/√(31)
Upper Limit = 76.707
98% Confidence interval is ( 69.293 , 76.707 ).