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.

Question

MathGuru · Answer

Use a one-sample z-test.

Hypotheses:
Ho: µ = 95 -->null hypothesis
Ha: µ < 95 -->alternate hypothesis

Formula:

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.

stylists · Answer

Need step by step format can't figure out, need an example:
two samples of data regarding salary of Co A and Co B determine if you can be 99% sure that Co?
A makes more than Co B? Company A - Mean 55,000, Std Dev 5,000, Sample Size 200; Co B mean 52,000, Std Dev 3,000 Sample size 400 please show steps thanks

patricia · Answer

two hundred math 115 students kept trak of the number of hours of homework they did during the fall 94 semester. the results are found to closely approximate a normal distribution with mean 14 hours and standard deviation 2 hours. determine the fraction of the students who studied more than 16 hours