IDS 270 Final Exam

UIN:______________________________

Please write your name on the top of the back of the very last page of what you turn in.

You MUST show your work for credit.

There are multiple version of the exam being used during this examination period. If
your exam shows work that is based on information from a different version of the exam

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then you will receive a zero for the entire exam. Initial to acknowledge: __________

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1. Define each of the following completely (including reference to the sample vs. the
population where appropriate). 5 Points
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mu-- population mean [1]

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s -- sample standard deviation [1]

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sigma-- population standard deviation [1]

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x-bar-- sample mean/average [1]

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sampling distribution-- a set of sample means from the same population [1]

[Note: Book has longer answer along the lines of ‘set of all possible statistics from
repeated samples blah blah blah’ That answer is also acceptable.

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 2. (10 pts.) A random sample of 36 postal workers is taken and their number of years on
the job is recorded. The calculated average number of years is 8 and the standard
deviation calculated from the sample is 4 (s=4). Find a 90% confidence interval for the
population mean number of years on the job for postal workers. Show all work.

d.f.=n-1=36-1=35,
No 35 on the table, so must use 30 for degrees of freedom
C.L. = 90% ==> t* = 1.697 [3']

m.e. = t* s/sqrt(n) = 1.697*4/6 = 1.131 [4']

[3']

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8 +/- 1.131 = (6.869, 9.131)

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 3. (4 pts.) Could you have used the results of the previous problem to test the following
hypothesis at alpha=0.10:

Ho: mu=7.5 vs.

Ha: mu not equal 7.5

If so, explain why and provide the answer. If not, explain why not.
This is a concept question. Do not perform any additional calculations.

[1] Yes
[2] Because 1-alpha=confidence and 2-sided test
[1] fail to reject H0 at level of 10%, because 7.5 is inside of the 90% C.I.

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 4. (15 points) A sample of 100 flights for United Airlines shows that 64 are on time. A
sample of 100 flights for American shows that 80 were on time. Using these figures
execute a hypothesis test at alpha=0.05 that there is no difference in the performance of
the two airlines.

That is, test the hypothesis that the proportion of on-time flights is the same for both
airlines vs. that there is a difference in on-time proportion.

Specify Ho and Ha, show the p-value, state whether you reject Ho or not and state your
general conclusion.

H0: p1=p2 [1]

Ha: p1≠p2 [1]

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p1-hat = 64/100 = 0.64 [1]

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p2-hat = 80/100 = 0.8 [1]

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p-hat = (64+80)/(100+100) = 0.72 [1]
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zob = (0.64-0.8) / sqrt( 0.72* (1-0.72)*( 1/100 +1/100)) [4]
= -2.52
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P-value = 2 * Pr (Z > |zob|) [4]

= 2* (1 - Pr (Z < 2.52))
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= 2 * (1- 0.9941)
=0.0118
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P-value < 0.05 ==> reject H0 for Ha at level of 5%. [1]

That is, there is strong evidence that there is some difference in the performance of the
two airlines. [1]
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 5. (15 points) A sample of 100 flights for United Airlines shows that 64 are on time. A
sample of 100 flights for American shows that 80 were on time. All flights that were not
on time are considered delayed. Analyze the data for no relationship between airline and
on time performance using a contingency table (Chapter 9, chi-square methods) at
alpha=0.10. Note this alpha is different from the one in the previous question.

Specify Ho and Ha, show the p-value, state whether you reject Ho or not and state your
general conclusion.

H0: there is No association between airline and on time performance [1]

Ha: there is some association between airline and on time performance [1]

Observed [2]
UA AA Total

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On time 64 80 144

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Delay 36 20 56

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Total 100 100 200

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Expected
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[2]
AA
On time 100*144/200=72 100*144/200=72
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Delay 100*56/200=28 100*56/200=28

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Chi-square [2]
UA AA
On time (64-72)^2/72=0.889 (80-72)^2 /72=0.889
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Delay (36-28)^2/28=2.286 (20-28)^2/28=2.286

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Total-Chisqr = 0.889 + 0.889 + 2.286 + 2.286 = 6.35 [1]

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d.f. = (2-1) (2-1) = 1 [1]

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p-value =Pr(X2 df=1 >6.35) belongs to (0.01, 0.02) [3]

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p-value < 0.10 ==> reject H0 for Ha at level of 10%. [1]

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That is, there is strong evidence that some association exists between airline and on time
performance. [1]
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 6. (25 points total) A group of researchers wants to know if a certain exercise program
increases bone density. They measure the bone density of 25 subjects. The subjects then
perform an exercise program for 6 months. The researchers know in advance that the
standard deviation of bone densities in the population is 2.0 (sigma=2.0). The bone
density for the sample of 25 subjects increased by an average of 0.6 points after the
exercise program.

Execute a hypothesis test at alpha=0.05 that the exercise program increased bone density
(mu is zero vs. mu is greater then zero). State Ho, Ha, calculate the appropriate statistic
and state your conclusion.

H0: mu =0 [1]
Ha: mu > 0 [1]

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zob=(0.6-0)/ (2/sqrt(25)) = 1.5 [2]

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p-value = Pr( Z>1.5) = 1- Pr(Z<1.5) = 1- 0.9332 = 0.0668 [4]

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p-value > 0.05 ==> fail to reject H0 at level of 5%. [1]
That is, there is no significant increase in the bone density. [1]
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What is the smallest increase that will allow you to reject Ho if mu is in fact 0?
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z_0.95 = 1.64 (or 1.65) [1]

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(xbar - 0)/ (2/ sqrt(25)) ≥ 1.64 (or 1.65)

==> xbar ≥ 1.64* 2/5+ 0 = 0.656 ( or 1.65*2/5 + 0 = 0.66) [4]
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What is the probability of rejecting Ho if bone densities increased in the population by

1.0 points (if mu was 1.0)? That is, what is the probability of getting the number you
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calculated above, or a larger number, if mu=1.0.

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z=(0.66 - 1) / (2/sqrt(25)) = -0.85 [2]

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power = Pr(Z> -0.85) = 1- Pr(Z<-0.85) = 1- 0.1977 = 0.8023 [4]

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Fill in the probabilities in the table below. Also indicate alpha, beta and power.
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What test concludes Ho is true in population Ho not true in population

Reject Ho Alpha=0.05 [1] Power=0.8023 [1]

Not reject Ho 0.95 [1] Beta=1-0.8023=0.1977 [1]

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 7. (3 points) If the population is U shaped, as shown in the graph below, provide a rough
sketch of the distribution of the averages if we took 2 sets of 50 samples. One of the sets
of samples has an n of 49 for each sample and the other has an n of 100 for each sample.
Label or provide a legend to distinguish between the two sampling distributions.
[1] point for bell shape for one sampling distribution. [1] point for bell shape for other
sampling distribution. [1] point for sampling distribution with larger n is tighter.

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8. (4 points) A recent article in the Make-Believe Journal of Social Science states that “A
large scale study concluded that teen-agers who watch an average of 3 hours or more of
television per day have lower grades”.
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The specifics of the test were:

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n=10,000 teen-agers in each group (those watching 3 hours or more vs. those watching
less than 3 hours).
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P-value=0.008
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Average difference in grades=0.002 on a 5-point scale.

Indicate if the results are practically vs. statistically significant below and explain why.
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Circle one: Statistically significant Not statistically significant

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Why?
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[1]Circle statistically significant. [1] Because p-value is smaller than any standard alpha
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Circle one: Practically significant Not practically significant

Why?
[1]Circle not practically significant. [1] Because 0.002 on a 5 point scale does not make a
difference in day to day life

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 9 (15 points) Bob looks in the classified ads for rents of one-bedroom and two-bedroom
apartments. He writes down the figures for 10 apartments of each type. He then
summarizes the two distributions as follows:

One bedroom: x-bar=531 and s=82.8

Two bedroom: x-bar=609 and s=89.3

Perform a hypothesis test that the average rent for a two-bedroom apartment is greater
than that for a one-bedroom apartment. Use alpha=0.10. State your null and alternative
hypotheses, calculate the test statistic, find the p-value and state your conclusion.

H0: mu (2br) = mu(1br) [2]

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Ha: mu(2br) > mu(1br) [2]

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tob= (609 - 531) / sqrt( 89.3^2/10 + 82.8^2/10) [3]

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= 2.03

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d.f. = min(10-1, 10-1)=9 [2]

p-value = Pr( T9 > 2.03) belongs to (0.025, 0.05) [4]

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p-value < 0.05 ==> reject H0 for Ha at level of 5%. [1]

That is, there is evidence that the rent for two-br apt is significantly greater than that for
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one-br apt. [1]

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 10. (2 points) We expect that as people invest more in stocks then they will have less
money to invest in bonds. If we execute a regression to predict the average amount
invested in bonds (the dependent variable, or Y) using the amount invested in stocks (the
independent variable, or X) then what would be the appropriate Ho and Ha regarding B1.

Ho: B1=0 [1]

Ha: B1<0 [1]

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(2 points) The standard deviation of Y is 100. The square root of the squared residuals

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divided by n-2 equals 50. Does this suggest that the regression fits the data well or that it

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does not fit the data well? Circle one.

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Fits the data well. Does not fit the data well.
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[2] Circle fits the data well.

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This
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