Question

A local newscaster reports that the average rainfall in the month of June is approximately 88 mm. However, a meteorologist wishes to test this claim, believing that the average rainfall in June is actually higher than 88 mm. He collects data on the average June rainfall for 10 randomly selected years, and computes a mean of 90 mm. Assuming that the population standard deviation is known to be 18.9, and that rainfall is normally distributed, determine each of the following:

a) What are the appropriate hypotheses:

		H0:μ=88,HA:μ≠88
		H0:X¯=88,HA:X¯≠88
		H0:X¯=88,HA:X¯>88
		H0:μ=88,HA:μ>88

b) Calculate the appropriate test statistic.

Round your answer to at least 3 decimal places.

c) What is the appropriate conclusion that can be made, at the 5% level of significance

		There is insufficient evidence to reject the null hypothesis, and therefore no significant evidence that the mean rainfall in June is different from 88 mm.
		There is sufficient evidence to reject the null hypothesis, and therefore conclude that the mean rainfall in June is more than 88 mm.

Answer 1

Solution:

a)

H0:μ = 88 , HA: μ > 88

b)

Test statistic z = = [90 - 88]/[18.9/10] = 0.33

Test statistic z = 0.335

c)

> sign in HA indicates that the test is RIGHT TAILED TEST.

p value = P[Z > z] = P[Z > 0.335] = 0.3688

Given level of significance is 5% i.e. 0.05

p value > 0.05

Fail to reject H₀

Conclusion is

There is insufficient evidence to reject the null hypothesis, and therefore no significant evidence that the mean rainfall in June is higher than  88 mm.