Question

A medical investigation claims that the average number of contaminations per week at the Main Hospital is 19.2. A random sample of 10 weeks had a mean number of 20.5. The sample standard deviation is 3.5.

Is there enough evidence to reject the investigator’s claim at alpha =.05?

Steps to be covered:

1. state the hypothesis and identify the claim
2. find the critical value from the table and mention the acceptance range
3. compute the test value (t test)
4. make the decision to reject or not the null hypothesis

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 19.2

H_a :    19.2

= 20.5

= 19.2

s = 3.25

n = 10

df = 10 - 1 = 9

Test statistic = t

= ( - ) / s / n

= (20.5 - 19.2) / 3.5 / 10

= 1.745

Test statistic = 1.745

P-value = 0.1149

= 0.05

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence to reject the investigator’s claim at alpha =0.05 .