Answers: Testing Part 1
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Problems
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 tvalue) 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 tvalue calculator and enter 24
degrees of freedom and a tvalue 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?
 H_{0}:
H_{A}:
 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) H_{0}: µ =
1550; H_{A}: 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
tvalue 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 onetailed 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 tvalue 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. (2tailed): .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 tvalue 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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