HYPOTHESIS TESTING

A researcher wishes to see if the mean number of days that a basic, low-price, small automobile sits on a dealer’s lot is 29. A sample of 30 automobile dealers has a mean of 30.1 days for basic, low-price, small automobiles. At α = 0.05, test the claim that the mean time is greater than 29 days. The standard deviation of the population is 3.8 days.

given X = 30.1 ( Sample Mean ) n = 30 ( Sample Size) μ = 29 ( Hypothesized Population Mean ) σ = 3.8 ( Population Standard Deviation ) α = 0.05 ( Level of Significance )

Formulate H0 and H1

A researcher wishes to see if the mean number of days that a basic, low-price, small automobile sits on a dealer’s lot is 29. A sample of 30 automobile dealers has a mean of 30.1 days for basic, low-price, small automobiles. At α = 0.05, test the claim that the mean time is greater than 29 days. The standard deviation of the population is 3.8 days.

H0: μ = 29 and H1: μ > 29

Specify the level of significance

α = 0.05

Select the appropriate test statistic

Establish critical region/regions

α = 0.05 (level of significance) and the test is a one- tailed test

Critical value Z = +1.65

Compute the actual value of the test statistic from the sample

Z = 30.1 – 29 3.8/ √ 30

= 1.59

Make a statistical decision

1.59 <1.65, and is not in the critical region, the decision is not to reject the null hypothesis.

z > zα

ConclusionThere is not enough evidence to support the claim that the mean time is greater than 29 days.

Problem 2:the average weight of 100 randomly selected sacks of rice is 48.54 kilos with a standard deviation of 20 kilos. Test the hypothesis at a 0.01 level of significance that the true mean weight is less than 50 kilos.

Given: X = 48.54 ( Sample Mean ) n = 100 ( Sample Size ) μ = 50 ( Hypothesized Population Mean ) s = 20 ( Standard Deviation ) α = 0.01 ( Level of Significance )

Step 1: H0: μ = 50 and H1: μ < 50Step 2: significance level α = 0.01Step 3: test statistic:

Step 4: critical regions: z < -zα which is z < -z0.01

Thus, we reject H0 if z < -2.33, otherwise we fail to reject H0

Step 5: Z = 48.54 – 50 20/ √100= -0.73

Step 6: statistical decision:

z < -zα

- 0.73 > -2.33Since z = - 0.73 is not in the critical region, the null hypothesis is not rejected.

Step 7: conclusion:The test result does not provide sufficient evidence to indicate that the true mean weight sack of rice is less than 50 kilos.