Question

Zingo's Grocery store claims that customers spend an average of 5 minutes waiting for service at the stores deli counter. A random sample of 45 customers was timed at the deli counter, and the average service time was found to be 5.75 minutes. Historically the standard deviation for waiting time is 1.75 minutes per customer.

a. Find the 95% confidence interval for the mean time that customers wait at the deli counter.

b. Is there sufficient evidence to indicate that the mean time that a customer spends waiting at the deli counter is 5 minutes? Use α = .02 significance level.

I am confused about whether to use a t value or a z value. Is the standard deviation given the population standard deviation or the sample standard deviation. The wording is very confusing to me. I am confused because it says random sample.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 5.75

sample standard deviation = s = 1.75

sample size = n = 45

Degrees of freedom = df = n - 1 = 45- =44

a) At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

t/2,df = t0.025,44 = 2.015

t /2,df = 2.015

Margin of error = E = t/2,df * (s /n)

= 2.015 * (1.75 / 45)

Margin of error = E = 0.5258

The 95% confidence interval estimate of the population mean is,

- E < <  + E

5.75 -0.5258< < 5.75+0.5258

5.224 < < 6.276

(5.224,6.276)

b)

= 5

This is the two tailed test .

The null and alternative hypothesis is

H0 :   = 5

Ha : 5

Test statistic = t

= ( - ) / s / n

= (5.75 - 5) / 1.75 / 45

= 2.875

p(t > 2.875 ) = 1-P (t< 2.875) = 0.0062

P-value = 0.0062

= 0.02

0.0062 < 0.02

Reject the null hypothesis .

There is sufficient evidence to suggest that the deli counter is 5 minutes.