Stock Prices are assumed to be affected by the annual rate of return. A sample of 28 stock prices along with
their annual rate of return was collected. The following resulted
INTERCEP
X1
Coefficient
-8.0
12.5
Standard
Error
3.0
1.3
T-test
P-value
2.67
9.61
0.0166
0.0001
1. Can you conclude that stock prices are linearly related to their annual rate of return?
2. What is the formula for the 95% confidence interval of the slope?
==========================
A fire insurance company wants to relate the amount of fire damage (y) in major residential fires ($000) to the
distance between the residence and the nearest fire station (x). The study is to be conducted in a large suburb
of a major city and a sample of 15 recent fires in this suburb is selected:
INTERCEPT
X
Coefficient
10.3
4.9
Standard T-test
Error
1.4
7.36
0.4
12.25
P-value
0.0001
0.0001
Lower 95% Upper 95%
4.03584
5.76416
3. Can you conclude that fire damage is linearly related to the distance to the station?
4. What is the 95% confidence interval of the slope?
==========================
. An experimenter wishes to compare the sales of two restaurants. She randomly selects eleven days from last
year for both restaurants and records the sales of each restaurant on those days. The output from Excel is
below:
t-Test: Two-Sample Assuming Equal Variances
Restaurant
1
2
Mean
30
55
Variance
353.6552 1657.742
Observations
11
11
Pooled Variance
1000
t Stat
-2.03541
P(T<=t) two-tail
0.055277
t Critical two-tail
2.085963
5. What is being tested in the above analysis?
6. What do you conclude from the above analysis?
Questions 7-14 are concerned with the following situation. A manager wishes to try three new training methods
to determine which will best increase the output of workers (in parts per minute). 15 workers are randomly
assigned to the three training methods (five to each method) and following the training their output is measured
.
6. Is this an experiment or observational study?
7. The type of experimental design being used is a __________.
9. What is the rejection region for testing that the mean output is the same for all training methods?
10. You are given that the MSA = 80 and MSW = 12. What is your conclusion?
11. Given the sample means for training method 1 was 14.5, method 2 was 20.1 and method 2 was 60, what
can you conclude?
12. In question 10 what was your null and alternative hypothesis?
13. In question 10 write a type I and a type II error in non-statistical language.
14. If you are performing a two-sided t-test which has 14 degrees for freedom, draw and indicate the rejection
region.
15. What is the general concept of hypothesis testing in analysis of variance?
Answers (Disclaimer: I am 95% sure the answers are correct).
1. Stock prices are linearly related to the annual rate of return since the p-value of 0.0001 is less than 0.05
2. A 95% confidence would use a t-value with 26 degrees of freedom = 2.0555
The interval is the “estimate ± t-table value * standard error” or 12.5 ± 2.0555 * 1.3
3. You can conclude that fire damage is linearly related to the distance from the fire station since the p-value is
less than 0.05. You can also say this because the interval for the slope does not contain zero
4. With 95% confidence we can say that for every mile a house is from the fire station, the average fire damage
will increase from $4,035.84 to $5,764.16
5. You are testing the null hypothesis that the mean sales of restaurant 1 is the same as the mean sales of
restaurant two for all days last year.
6. The 95% confidence interval uses a t with 20 degrees of freedom = 2.086.
Since
1 1
(30  55)  2.0860 1000(  )  25  28.13
11 11
Contains the value 0, we can conclude there is no significant difference between the average sales of the two
restaurants. (or we can say that since the p-value is not less than 0.05 we can not say the average sales are
different)
7. This is an experiment because similar workers are being assigned to the teaching methods.
8. This is a completely randomized design.
9. c = 3 and n = 15. the F value would then have 2 and 12 degrees of freedom = 3.89 assuming  = 0.05
You would reject the null hypothesis if the F > 3.89
10. The F test value is 80/12 = 6.67. Since it is larger than 3.89 we can conclude that the average output is not
the same for all training methods.
11. Using Tukeys procedure we would use a Q table value of 3.77. The margin of error (or critical range) is
Q
MSW
2
1 1
12  1 1 
    3.77
    5.84
n
nj
2
5 5
 i

The rule we would use is that any difference in sample means that exceeds 5.84 would allow us to conclude the
population means are different.
X 1  X 2  14.5  20.1  5.6 , X 1  X 3  14.5  60  45.5 , X 2  X 3  20.1  60  39.9
We can say that teaching 3 is better than 2 and 3 is better than 1 but we can not tell the difference in 1 and 2.
12. Null hypothesis = all means are equal 1   2   3
Alterative hypothesis = not all means are equal.
13. A type 1 error would be saying there is a difference in the average output among the teaching methods
when actually there is no difference.
A type 2 error would be saying there is no significant differences in the average output among the teaching
methods when actually there are differences.
14. The t-table value with 14 degrees of freedom is 2.1448. You would then reject the null hypothesis if the ttest value was larger than 2.1448 of smaller than -2.1448
15. Sample means vary. Analysis of Variance determines if the sample means vary enough to say that the
population means are different. It does this by comparing the variability of sample means to the variability
within the samples.