Statistics
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 onesample ztest.
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 ztable at .05 level of significance for a onetailed test.
If the ztest statistic exceeds the critical value from the ztable, reject the null. If the ztest statistic does not exceed the critical value from the ztable, 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:
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 
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