Question

You were told that the average number of calories in a low-calorie meal is 450.   Your friend started making low calorie meals at her restaurant.  You took a sample of 27 such meals and the mean for these meals is 440 and the standard deviation is 22 calories.   You want to test the hypothesis that the average for this sample is different from the national average, your friend told you that she is sure that her meals have fewer calories.  Do the test again.  Use α=.05.

a. what is the alternative hypothesis?

b. what is the rejection region?

c. should the null hypothesis be rejected?

d. calculate the 95% confidence interval for the population mean.

Answer 1

Part a)

To Test :-
H0 :- µ = 450
H1 :- µ ≠ 450

Part b)

Reject null hypothesis if | t | > t(α/2, n-1)
Critical value t(α/2, n-1) = t(0.05 /2, 27-1) = ± 2.056

Reject null hypothesis if t > 2.056 OR t < -2.056

Part c)

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 440 - 450 ) / ( 22 / √(27) )
t = -2.3619

| t | > t(α/2, n-1) = 2.3619 > 2.056
Result :- Reject null hypothesis

Part d)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 27- 1 ) = 2.056 ( Critical value from t table )
440 ± t(0.05/2, 27 -1) * 22/√(27)
Lower Limit = 440 - t(0.05/2, 27 -1) 22/√(27)
Lower Limit = 431.2951
Upper Limit = 440 + t(0.05/2, 27 -1) 22/√(27)
Upper Limit = 448.7049
95% Confidence interval is ( 431.2951 , 448.7049 )