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 sixteen. 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?

Answer 1

Answer :

we have given :

n  = sample size  = 16

x̄ = sample mean  = 13

μ = population mean = 16

s = sample standard deviation  = 5

α = 1 % = 0.01

## step1 : null and alternative Hypothesis :

Ho : μ = 16 vs H1: μ < 16

( it is one tailed test )

## step 2 : Test statistics :

t = (x̄ - μ) * sqrt ( n ) / s

t   = ( 13 - 16 ) * sqrt ( 16)   / 5

= -12 / 5

= - 2.4

## step 3 :  level of significance  = 0.01

## step 4 :  P value : P (t < - 2.4 )

df = degree of fredom = n -1 = 15

= 0.0149 ( use statistical table )

step 5 : Decision  :

We reject Ho if p value is less than alpha value using p value approach here
p value is greater  than alpha value we fail to reject Ho at given level of significance .

Step 6 : Conclusion :

There is Inufficient evidence to conclude that population mean is less than 16 hours .

ie we can say that population mean is equal to 16 hours .