Question

In: Statistics and Probability

The post anesthesia care area (recovery room) at St. Luke's Hospital in Maumee, Ohio, was recently...

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)

Solutions

Expert Solution

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   


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