Question

An SRS of 16 Spokane County Schools' seniors had a mean SAT Verbal score of 500 with a standard deviation of s = 100. We know that the population is normally distributed. We wish to determine a 90% confidence interval for the mean SAT Verbal score µ for the population of all seniors in the district.

a. Using the above data, the critical value has how many degrees of freedom?

b. What is the critical value for the 90% confidence interval?

c. Find the 90% confidence interval.

d. Suppose the population of all high school seniors has a mean score of 450. We wish to see if the data provide evidence that the mean score of seniors is larger than 450. Write the hypotheses for this test.

e. Carry out the appropriate test of significance at the 5% level and write your conclusions.

Answer 1

a)

df = n - 1 = 16 - 1 = 15

b)

t critical value at 0.10 significance level with 15 df = 1.753

c)

90% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
500 ± 1.753 * 100/√(16)
Lower Limit = 500 - 1.753 * 100/√(16)
Lower Limit = 456.175
Upper Limit = 500 + 1.753 * 100/√(16)
Upper Limit = 543.825
90% Confidence interval is ( 456.175 , 543.825 )

d)

H0: = 450

Ha: > 450

e)

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 500 - 450 ) / ( 100 / √(16) )
t = 2

Test Criteria :-
Reject null hypothesis if t > t(α, n-1)
Critical value t(α, n-1) = t(0.05 , 16-1) = 1.753

t > t(α, n-1) = 2 > 1.753

Result :- Reject null hypothesis

we have sufficient evidence to conclude that the mean score of seniors is larger than 450.