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1. The mayor of a large city claims that the average age of persons arrested is 22.3 years. An investigative reporter doubts this claim, believing that the average age is younger. He takes a sample of 100 cases and finds that average age a bit less than 22.3. Is this conclusive proof that the mayor is a liar? Explain. (No need for any computation.)

No. Because there will be some random variation from sample to sample, it is possible for the claim to be correct and the sample mean to be less than the claimed mean. However, if that difference is great enough, it becomes very unlikely that a sample like this would be picked by random chance, so we start doubting that the difference is due to random chance and start suspecting it is due to a bad claim.

2. Suppose that a library finds, by utilizing its computer records, that the mean number of pages in its books is 300 with a standard deviation of 60. If a student doing an assignment for a statistics class takes a random sample of 36 books, what is the probability of getting a sample mean that is more than 15 pages away from the true mean?

The standard error of the mean is 60/6 = 10. So we expect the sample mean to be 300 give or take 10. An error of 15 is 1.5 standard errors above the mean, and using the Normal tables, we find that the probability of being 1.5 standard errors above the mean is .0668.

3. Suppose that a credit card company finds, using its computer records, that the average customer owes \$750 with a standard deviation of \$250. If a random sample of 225 customers is taken, what is the probability of getting a sample mean that is more than \$25 away from the true mean?

The standard error of the mean is \$250/15 = \$50/3 = \$16.67. So we expect the sample mean to be \$750 give or take \$16.67. An error of 25 is 1.5 standard errors above the mean, and using the Normal tables, we find that the probability of being 1.5 standard errors above the mean is .0668.

4. One area in which industry uses hypothesis testing is in quality control. For example, suppose that a manufacturer of candy bars wants its bars to weigh an average of 2 ounces. Because of wear, tear, and vibration, the machinery making the bars will occasionally make the bars too small or too big and will need adjustment. To see if the machinery is working satisfactorily, 25 bars are periodically checked and their average weight found.

a) Formulate the hypothesis and the alternative and state a decision rule (in terms of a t-value) if alpha = .05.
b) Suppose that the sample average is 2.036 ounces and the sample standard deviation is 0.090 ounces. How many standard errors is the sample mean away from the claimed mean?
c) How likely is this difference if it is due to random chance?
d) Should we accept random chance as the explanation of the difference? What action should the company take? Explain.

a) The claim is that the machine is working correctly, or that the population mean is 2 oz. The alternative is that the population mean is not 2 oz. With 24 degrees of freedom, we will accept the claim if t is greater than -2.06 and less than 2.06.
b) The standard error for the sample mean is 0.090 divided by the square root of n, the sample size. This is a measure of how much we expect the sample mean to vary from sample to sample. 0.090 ÷ 5 = 0.45. Hence, the sample mean is two standard errors away from the claimed mean.
c and d)) If we check a t-value calculator and enter 24 degrees of freedom and a t-value of 2, we find that the probability of being that far or further from where we expect to be is .0569. That is enough for us to attribute this difference to random chance rather than to the claim being wrong. Hence, we should do nothing.

5. A manufacturer of light bulbs wants to know whether its lightbulbs are better or worse than the lightbulbs of a competitor. The company's statistician knows that the company's lightbulbs last an average of 1550 hours.

a) What is the null hypothesis and what is the alternative?
H0:
HA:
b) Taking a random sample of 100 of the competitor's lightbulbs, he finds a mean lifetime of 1585 hours with a standard deviation of 150 hours. If alpha = .05, should he reject the null hypothesis? Why or why not?

a) H0: µ = 1550; HA: µ ≠ 1550
b) If the claim is correct, we expect the sample mean to be 1550 give or take 15 hours. The 15 hours comes from dividing the sample standard deviation by the square root of the sample size (100). The sample mean is 2.33 standard errors above the claim, and getting a sample mean this far from the claim should happen only 2.17% of the time according to the t-value calculator. This is too unlikely for us to attribute the difference to random chance. Instead we reject the claim and decide the evidence favors the alternative.

6. A box contains four envelopes. One envelope has a \$50 bill, two envelopes have \$10 bills, and one has a blank piece of paper.

a) If you randomly pick an envelope from this box, what is the expected value of your pick?
b) Suppose you have to pay \$1.00 to pick and envelope from this box. You pay the \$1.00 and the envelope you pick has a blank piece of paper. Did you make a mistake in playing this game? Explain carefully.

a) The expect value is \$70 ÷ 4 = \$17.50.
b) You did not make a mistake unless that dollar you lost was really, really important to you. Life is full of small risks, and you should take those that are in your favor, as this one is. Similarly, the procedures of hypothesis testing will sometimes give you erroneous decisions, but that does not mean that you made a mistake in using the procedure. The alternative is not using the procedure, which can be far worse.

7. Suppose a consumer group wants to test the claim that the average number of raisins in a brand of cereal is at least 300 per box. It takes a random sample of 100 boxes, counts the raisins, and finds that the standard deviation is 20 raisins. If they use alpha at .05, what range of sample means will convince them that the company's claim is inaccurate? Explain very carefully how you get your answer.

This is a one-tailed test because we will only find the company at fault if it has too few raisins, not too many. The standard error of the mean, that is, how much we expect the sample mean to vary from sample to sample, is 20/10 = 2. Using a t-value calculator with 99 degrees of freedom (sample size less one), we find that only 5% of the time will the sample mean be 1.66 standard errors below the true mean. Multiplying 1.66 by 2 we get 3.32. Hence, if the sample mean is less than 300oz - 3.32 oz = 296.68 oz, we will reject the claim of the company.

8. Suppose a consumer group doubts the claim that the average number of raisins in a brand of cereal is at least 300 per box. They take a sample of 14 and get the result below. What value belongs under t?

N: 14
Mean: 298.6429
Std. Deviation: 18.2234
Std. Error Mean: 4.8704
Test Value = 300
t: ____
df: ____
Sig. (2-tailed): .785
Mean Difference: -1.3571

What belongs in df?
What is the value of t?
What do they conclude?

Degrees of freedom are 13, one less than sample size of 14.
The t-value is the Mean Difference divided by the Standard Error of the Mean, or -1.3571/4.8704 = -.2786.
They conclude that the difference between the sample results and the claim is small enough so that random chance is a plausible explanation of the difference. The sample results are consistent with the claim being true.

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