Question

# Suppose the average number of calories in low-calorie dinner is 600 with a standard deviation of...

Suppose the average number of calories in low-calorie dinner is 600 with a standard deviation of 100 calories. Your friend sells healthy dinners at her restaurant. You believe that, in fact, the average calories for average dinner is different from 600.  To test the hypothesis, you took a sample of 16 dinners at your friend’s restaurant.  The average for this group is 650 calories.

a. what is the value of the test statistic?

b. should the t-test be used?

c. what is the rejection region

d. should the null hypothesis be rejected?

e. what is the 95% confidence interval for the population mean?

Part a)

Test Statistic :-
Z = ( X̅ - µ ) / ( σ / √(n))
Z = ( 650 - 600 ) / ( 100 / √( 16 ))
Z = 2

Part b)

No, since the population standard is known.

Part c)

Reject null hypothesis if | Z | > Z( α/2 )
Critical value Z(α/2) = Z( 0.05 /2 ) = 1.96
Reject null hypothesis if Z > 1.96 OR Z < - 1.96

Part d)

| Z | > Z( α/2 ) = 2 > 1.96
Result :- Reject null hypothesis

Yes, null hypothesis is rejected.

Part e)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96 ( Critical value from Z table )
650 ± Z (0.05/2 ) * 100/√(16)
Lower Limit = 650 - Z(0.05/2) 100/√(16)
Lower Limit = 601
Upper Limit = 650 + Z(0.05/2) 100/√(16)
Upper Limit = 699
95% Confidence interval is ( 601 , 699 )

#### Earn Coins

Coins can be redeemed for fabulous gifts.