Question

: A sample of 9 days over the past six months showed that a dentist treated the following numbers of patients: 22, 25, 20, 18, 15, 22, 24, 19, and 26. If the number of patients seen per day is normally distributed, would an analysis of these sample data reject the hypothesis that the variance in the number of patients seen per day is equal to 10?

Use α .10 level of significance. What is your conclusion?

Answer 1

Solution:

We are given a data of sample size n = 9

22, 25, 20, 18, 15, 22, 24, 19, 26

Using this, first we find sample mean() and sample standard deviation(s).

=   

= (22 + 25.......+ 26)/9

= 21.22

Now ,

s =   

s = (22 - 21.22)2 + (25 - 21.22)2 + ......+ (26 - 21.22)2

s = 3.56

Using given data, find Xi - for each term.Take square for each.Then we can easily find s.

s = 3.56

Solution:

1)

The null and alternative hypothesis are

H0 : = 10  

Ha :    10

2)

The test statistic t is

t =   = [21.22 - 10]/[3.56 /9] = 9.455

The value of the test​ statistic t = 9.455

3)

Now ,

d.f. = n - 1 = 9 - 1 = 8

sign in Ha indicates that the test is TWO TAILED.

t = 9.455

So , using calculator ,

p value = 0.000

4)

p value is less than the significance level 0.10.

Decision: Reject the null hypothesis H0

5)

Conclusion : There is sufficient evidence to support the claim that the average percent cover has changed.