Question

After implementing a new strategy to improve services for patients, the Emergency Department of a hospital claims customers spend fewer than 20 minutes on average to see a physician. A randomly selected sample of 132 patients shows the average wait time of 18.2 minutes. Test the Emergency Department's claim at the 10% level of significance, assuming the standard deviation of wait time for all patients in that hospital is 2.3 minutes.

Answer 1

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: Customers spend 20 minutes on average to see a physician.

Alternative hypothesis: Ha: Customers spend fewer than 20 minutes on average to see a physician.

H0: µ = 20 versus Ha: µ < 20

This is a lower tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 20

Xbar = 18.2

σ = 2.3

n = 132

α = 0.10

Critical value = -1.2816

(by using z-table or excel)

Z = (18.2 - 20)/[2.3/sqrt(132)]

Z = -8.9915

P-value = 0.0000

(by using Z-table)

P-value < α = 0.10

So, we reject the null hypothesis

There is sufficient evidence to conclude that Customers spend fewer than 20 minutes on average to see a physician.