Question

The owner of a pizza restaurant in France knows that the time customers spend in the restaurant on Saturday evening has mean 90 minutes and standard deviation 15 minutes. He has read that pleasant odors can influence customers, so he spreads a lavender odor throughout the restaurant.

The average time spent by the 30 customers the next Saturday evening is 105.7 min  

1. Suppose that the standard deviation (15 minutes) is not changed by the odor. Is there a reason to think that the lavender odor has increased the mean time customers spend in the restaurant? Use statistics for significance test.

Answer 1

H0: = 90

Ha: > 90

Test Statistic :-
Z = ( X - µ ) / ( σ / √(n))
Z = ( 105.7 - 90 ) / ( 15 / √( 30 ))
Z = 5.73
Test Criteria :-
Reject null hypothesis if Z > Z(α)
Critical value Z(α) = Z(0.05) = 1.645
Z > Z(α) = 5.73 > 1.645
Result :- Reject null hypothesis

We have sufficient evidence to support the claim that the mean time customers spend in the restaurant

ahs increased.