Question

Chapter 10 - 32

The post anesthesia care area (recovery room) at St. Luke’s Hospital in Maumee, Ohio, was recently enlarged. The hope was that the change would increase the mean number of patients served per day to more than 25. A random sample of 15 days revealed the following numbers of patients.

25

27

25

26

25

28

28

27

24

26

25

29

25

27

24

Answer 1

Since we are comparing one sample mean with the population mean and the population standard deviation is not known, the t-test for One Population Mean is be used to test the hypothesis that the mean number of patients served per day is more than 25.

Hypothesis

The Null and Alternative Hypotheses are defined as,

This is a left tailed test.

Let the significance level = 0.05

Test-statistic

The t statistic is obtained using the formula,

For calculation purpose, the mean and standard deviation are obtained in excel using the function =Average() and =STDEV(). The screenshot is shown below,

Now,

P-value

The p-value is obtained from t distribution table for t = 2.6935 and degree of freedom = n - 1 = 15 - 1 = 14. (In excel use function =T.DIST.RT(2.6935,14))

Conclusion

Since the P-value is less than significance level = 0.05 at 5% significance level, the null hypothesis is rejected. Hence we can conclude that mean number of patients served per day is significantly more than 25.