The weights of the apples grown on an orchard are normally distributed. The mean weight has been m0 = 9.500 ounces. Because of the inclement weather this year, the farmer would like to know if the mean weight has decreased. A random sample of n = 16 apples is selected. The sample mean is x = 9.320 ounces and the sample standard deviation is s = 0.18 ounces. A test needs to be conducted to test if the population mean has decreased at a significance level a = 0.05.

23. The null and alternative hypotheses should be (3)

24. Using the action limits to set up the decision rule, the decision rule should be (3)

25. The test statistic of this test is (3)

26. The conclusion of this test should be (3)

t (s) is in rejection zone then we reject H₀.

Bad weather indeed make apples weight decrease

Step-by-step explanation:

Normal Distribution

population mean μ₀ = 9.5 ou

sample size = n = 16 then we should apply t-student table

degree of fredom df = n - 1 df = 16 - 1 df = 15

1.-Test hypothesis

H₀ null hypothesis μ₀ = 9.5

Hₐ alternative hypothesis μ₀ < 9.5

One left tail-test

2.-Confidence level 95 %

α = 0,05 and df = 15 from t-student table we get t (c) = - 1.761

3.-Compute t (s)

t (s) = [ μ - μ₀ ] / √s/n t (s) = (9.32 - 9.5) * √16 / 0.18

t (s) = - 0.18*√16 / 0.18

t (s) = - 4

4.-Compare t (s) and t (c)

t (s) < t (c) - 4 < - 1.761

Then t (s) is in the rejection zone.

5. - Decision

t (s) is in rejection zone then we reject H₀.

Farmer conclude that bad weather make apples weight decrease