Question

3.Every year, the Hometown Hardware Store expects an average family to spend at least $606.40 on springtime home repairs. A new analyst in the store disputes this and challenges that the average spending per family is less than this amount.

The analyst conducts a test on the basis of a random sample of 30 households, using a 5% significance level, resulting in a sample mean of $589.95. This is normally distributed using a population standard deviation of $65.

Ho: LaTeX: \mu\:\:\:

μ

>= 606.40 Ha:

Critical Value = 2 decimal places

Statistical Value = 3 decimal places

pvalue =

Decision: Reject or Do Not Reject

Answer 1

Ans. Let sample average expenditure, m = $589.95

Population standard deviation, s = $65

Sample size, n = 30

Standard error of average expenditure, se = s/n^0.5 = 11.87

Hypothesis,

H0: mu >= 606.40

H1: mu < 606.40 (lower tail test)

zstatistic = (m - mu)/se = (589.95 - 606.40)/11.87 = -1.3858

zcritical at level of significance 5% = -1.645

p-value = 0.0838 or 8.38%

As pvalue is greater than the level of significance, so, we fail to reject the null hypothesis. Hence, an average family spends at least $606.40.

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