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February 27, 2015

February 27, 2015

Posted by **Stuart** on Sunday, March 1, 2009 at 1:02pm.

a) The average college student spends no more than $300 per semester at the university's bookstore.

b) The average adult drinks 1.5 cups of coffee per day.

c) The average SAT score for entering freshmen is at least 1200.

d) The average employee put in 3.5 hours of overtime last week.

I have the answers for a) H0: u greater or equal to 300 H1: u < 300 so its a one tailed test. Draw the bell curve and shade the left side of it. That’s the rejection region

b) H0: u is = to 1.5

H1: u is not = to 1.5 this is the two tailed test. Reject the two sided of the bell curve but not the middle

Need the answer for C and D.

2) The International Coffee Association has reported the mean daily coffee consumption for U.S. residents as 1.65 cups. Assume that a sample of 38 people from a North Carolina city consumed a mean of 1.84 cups of coffee per day, with a standard deviation of 0.85 cups. In a two-tail test at the 0.05 level, could the residents of this city be said to be significantly different from their counterparts across the nation?

3) During 2002, college work study students earned a mean of $1252. Assume that a sample consisting of 45 of the work study students at a large unversity was found to have earned a mean of $1277 during that year, with a standard deviation of $210. Would a one-tail test at the 0.05 level syggest the average earnings of this university's work study students were significantly higher than the national mean?

- Statistics -
**MathGuru**, Sunday, March 1, 2009 at 2:26pmI'll give you a few hints for 1).

Rethink part a). The statement says "no more than $300" which means $300 or less. You are correct in that this one is a one-tailed test.

Part b) looks good!

For part c): The statement is says "at least 1200" which means 1200 or more. (This test will be one-tailed.)

Part d) is similar to part b).

For 2): Use a one-sample z-test.

Here's a formula:

z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)

With the data:

z = (1.84 - 1.65)/(0.85/√38) = ?

Finish the calculation. Determine the critical or cutoff value to reject the null by looking at a z-table using 0.05 for a two-tailed test. If the test statistic exceeds either critical value (in either direction), then the null is rejected in favor of the alternative hypothesis. You can then conclude there is a difference. If the test statistic does not exceed either critical value, then the null is not rejected and you cannot conclude a difference.

For 3): Use the same process with the same formula as 2). Remember that this one is a one-tailed test, but the procedures to determine whether or not to reject the null are the same.

I hope this will help.

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