Question

A sample of 9 days over the past six months showed that a dentist treated the following numbers of patients: 22, 24, 21, 19, 15, 21, 24, 18, and 27. 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 a .10 level of significance.

1. State the null and alternative hypotheses.
   
   ₀:  ² Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 1
   _a:  ² Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 2
   
2. Calculate the value of the test statistic (to 2 decimals).
   
   
3. The p-value is Selectless than .02between .02 and .05between .05 and .10between .10 and .20greater than .20Item 4
   
4. What is your conclusion?

Answer 1

a. H0: vs H1:

b.

​=(9−1)*12.94444​/10

=10.356

10.36

c. Here the degrees of freedom of is (9-1) = 8.

As the test is a two-tailed test so the P-value = 2*min{P[ > 10.36] or P[ < 10.36]}

So, here P[ > 10.36] = 0.240662 {Using R code pchisq(10.36,8,lower.tail = F)} and P[ < 10.36] = 0.759338. {Using R code pchisq(10.36,8,lower.tail = T)}

So, the P-value is 2*0.240662 = 0.4813241.

The P-value is greater than 0.2.

d. So, we can't reject the null hypothesis.

So, we can conclude at 20% level of significance that the population variance is 10.