posted by vera on .
A newspaper article stated that college students at a local university spent an average of $95 on beer a year. A student investigator who believed the average stated was too high polled a random sample of 50 students and found a mean of $92.25 and s = 10 Use these results to test at the .05 level of significance the statement made by the newspaper.
Use a one-sample z-test.
Ho: µ = 95 -->null hypothesis
Ha: µ < 95 -->alternate hypothesis
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (92.25 - 95)/(10/√50) = ?
Finish the calculation.
Check a z-table at .05 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null. If the z-test statistic does not exceed the critical value from the z-table, do not reject the null. If you reject the null, the conclusion will be µ < 95.
I hope this will help get you started.
Need step by step format can't figure out, need an example:
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