Question

A coffee shop owner is worried about a slowdown in business because a new shop opened up one block away. Before the new shop opened, the owner’s customer traffic (customers per hour) was normally distributed with a mean of fourteen. The owner randomly surveyed sixteen hours and recorded a mean of thirteen customers per hour, a standard deviation of five customers per hour. Develop null and alternative hypothesis that will help deciding whether his shop is in trouble (when slowdown in business occurs) at the 1% significance level and calculate the p value? Question options:

0.2181

0.0008

0.9992

0.9912

Answer 1

H0: = 14

Ha: < 14

Test statistics

t = - / (S / sqrt(n) )

= 13 - 14 / (5 / sqrt(16) )

= -0.8

This is test statistics value.

Degrees of freedom (df) = n - 1 = 16 - 1 = 15

From T table,

p-value at test statistics of -0.8 and df of 15 = 0.2181

Decision - Since p-value > 0.01 significance level, we fail to reject H0.