Question

No customer wants to wait a long time in a checkout line. A retailer established a policy for mean time to check out at 240 seconds. After opening a new location, the retailer asked an analyst to see weather the new store was exceeding this standard (weather customers were waiting too long to check out on average)

The analyst took a random sample of 30 customers at the new location and measured the time they had to wait to check out. In the sample of customers, the mean waiting time was 274 seconds. Drawing on past experience with checkout times, the analyst assumed a population standard deviation of 70 seconds.

Preform and interpret a hypothesis test to determine weather the population mean waiting time at the new location is longer than 240 seconds. use an α =0.01 significance level.

Answer 1

Given that, population standard deviation = 70 seconds

sample size (n) = 30 and sample mean = 274 seconds

The null and alternative hypotheses are,

H0 : μ = 240 seconds

Ha : μ > 240 seconds

This test is right-tailed.

Test statistic is,

=> Test statistic is, Z = 2.66

critical value at significance level of α = 0.01 is, Z^* = 2.33

Using z-table, we find the p-value,

p-value = P(Z > 2.66) = 1 - P(Z < 2.66) = 1 - 0.9961 = 0.0039

=> p-value = 0.0039

Since, p-value = 0.0039 < α = 0.01, we reject the null hypothesis.

Conclusion : There is sufficient evidence to conclude that, the population mean waiting time at the new location is longer than 240 seconds.