Question

The post anesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently enlarged. The hope was that with the enlargement the mean number of patients per day would be 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

At the .01 significance level, can we conclude that the mean number of patients per day is more than 25?

1. Identify Null and Alternate Hypothesis
2. Identify the t statistic
3. Estimate the p-value and interpret it. (Reject or Not Reject and why)

Answer 1

Solution:

x	x2
25	625
27	729
25	625
26	676
25	625
28	784
28	784
27	729
24	576
26	676
25	625
29	841
25	625
27	729
24	576
∑x=391	∑x2=10225

Mean ˉx=∑xn

=25+27+25+26+25+28+28+27+24+26+25+29+25+27+24/15

=391/15

=26.0667

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√10225-(391)21514

=√10225-10192.066714

=√32.933314

=√2.3524

=1.5337

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :    = 25

Ha : > 25

Test statistic = t

= ( - ) / S / n

= (26.07 - 25) / 1.53 / 15

= 2.709

Test statistic = t =  2.709

P-value =0.0085

= 0.01  

P-value <

0.0085 < 0.01

Reject the null hypothesis .

There is sufficient evidence to suggest that