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.
State the null hypothesis and the alternative hypothesis for testing that the average back-to-school spending differs from $606.40.
State the critical value(s) for testing this hypothesis.
Calculate the test statistic Z for testing this hypothesis
State and support your conclusions
n = 20
claim :The average back-to-school spending differs from 606.40.
Null and alternative hypothesis is
Level of significance = 0.05
The z-critical values for a two-tailed test, for a significance level of α=0.05 is 1.96
critical value = 1.96
Here population standard deviation is known so we use z-test statistic.
Test statistic is
Rejection Region : critical value Test statistic , Reject Ho
conclusion : There is sufficient evidence to support the claim that the average back-to-school spending differs from 606.40.
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