Question

# A research analyst disputes a trade group's prediction that back-to-school spending will average \$606.40 per family...

A research analyst disputes a trade group's prediction that back-to-school spending will average \$606.40 per family this year. She believes that average back-to-school spending will significantly differ from this amount. She decides to conduct a test on the basis of a random sample of 20 households with school-age children. She calculates the sample mean as \$628.85. She also believes that back-to-school spending is normally distributed with a population standard deviation of \$45. Use 5% significance.

Part 1

State the null hypothesis and the alternative hypothesis for testing that the average back-to-school spending differs from \$606.40.

Part 2

State the critical value(s) for testing this hypothesis.

Part 3

Calculate the test statistic Z for testing this hypothesis

Part 4

n = 20

claim :The average back-to-school spending differs from 606.40.

part 1

Null and alternative hypothesis is

Level of significance = 0.05

part 2

The z-critical values for a two-tailed test, for a significance level of α=0.05 is 1.96

critical value = 1.96

part 3

Here population standard deviation is known so we use z-test statistic.

Test statistic is

Rejection Region : critical value Test statistic , Reject Ho

Part 4

conclusion : There is sufficient evidence to support the claim that the average back-to-school spending differs from 606.40.

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