Question

At a large hospital average number of new acquired infection is normally distributed with u=17.5 and standard deviation of 2.5. After the new procedure were implemented, an SRS of 36 weeks found a mean 15 infection per week. At 5% level of significance (a=0.05), is this evidence that the mean infection is different from 17.5 infections per week.

a. state the hypotheses

b. calculate the value of the test statistic.

c. determine the p-value

d. state your conclusion in terms of the problem.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 17.5

H_a : 17.5

Test statistic = z

= ( - ) / / n

= (15 - 17.5) / 2.5 / 36

Test statistic = -6

P(z < -6) = 0

P-value = 2 * 0 = 0

= 0.05

P-value <

Reject the null hypothesis .

There is sufficient evidence to conclude that the mean infection is different from 17.5 infections per week.