Download Section 7.3

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Homework #7
Most of these are problems from your textbook. Remember to get the data for the problems that need it in the
lab (S:data files/IPS4e/SPSS etc.), off the web (http://bcs.whfreeman.com/ips4e/ click on Datasets and
choose Excel format since SPSS can read it. Note: you must have a zip program, like ZipCentral or
Winzip, to unzip the file), or off the cd in the back of your book (D:PCDataSets/SPSS/).
Section 7.1
**1. Ex. 7.2 & 7.4 In SPSS open the data file Ch07/EX07_002.por, Select Analyze  Compare Means  OneSample T Test  Options  Confidence Intervals 95%  Continue,move [rent] into  Test Variable(s): ,change
Test Value to “500”  Click “OK”
One-Sample Statistics
Normal Q-Q Plot of
700
N
Mean
Std.Deviation
Std.Error Mean
600
Expected Normal Value
10
531.0000
82.7916
26.1810
s
,
n
what is (531  2.262  26.1810, 531  2.262  26.1810) 
(471.7745,590.2255)
95% CI for  : x  t9,.025
500
400
300
300
400
500
600
700
Observed Value
One-Sample Test
Test Value = 0
t
df
20.282
Sig. (2-tailed)
9
.000
Mean
Difference
531.0000
95% Confidence Interval of the
difference
Lower
Upper
471.7745
590.2255
7.4 For the data, x = 531 and s=82.792. H0:  = $500 vs. Ha:  > $500. The t statistic is 1.184 with 9 df.
From software, the p-value is 0.1335, half of .267 the p-value given which is for a two-sided test. From
Table D, the t statistic is between the 0.15 and 0.10 columns. This is not strong evidence that the mean is
over $500.
One-Sample Test
Test Value = 500
t
1.184
t
df
9
Sig. (2-tailed)
.267
Mean
Difference
31.0000
95% Confidence
Int erval of the
Difference
Lower
Upper
-28.2255
90.2255
x  0
531  500

 1.184
s / n 82.792 / 10
NOTE: $500 is in the 95% confidence interval, so $500 is a plausible mean rent of all advertised
apartments.
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2. 7.6 (a) H0:  = $0 vs. Ha:   $0, where  is the mean change in sales. We use the two-sided
alternative because there is no information in the problem that suggests whether we are looking for an
increase or a decrease. (b) The t-statistic is 2.263 and the p-value is 0.028 from software. Table D does
not have 49 df, but for 50 df, our statistic is between 2.109 and 2.403, so we feel comfortable saying that
the (2 sided) p-value is between 0.04 and 0.02. (c) Certainly not. We are confident that the mean sales
are up, but individual stores may either increase or decrease.
Extra question: In part (b), would the appropriate Confidence Interval provide the same conclusion.
Perform the appropriate calculations to support your answer.
s
0.15
ANSWER: Yes. C.I. = x  t49,.025
 .048  2.009
 (0.535,9.065) . Since 0 is NOT in the
n
50
interval, we would conclude that average sales are different, but we don’t know whether that means
increase or decrease.
3. 7.12 (a) 19 (b) 1.729 and 2.093 (c) 0.05 and 0.025 (d) Since it is a one-sided test, 0.05 and 0.025. (e)
Is significant at 5%, since the statistics is larger than the table value (and our alternative is “greater
than”). Not significant at 1% because it is not larger than 2.093. (Omit part (f)but here’s the answer) (f)
0.0385.
**4. The power of a t test is the numerical measure (between 0 and 1) of the test’s ability to detect
deviations from the null hypothesis. Which of the following is NOT true (about the power of a t test?
(a) Sample size affects the power of the test we have used.
(b) Power of the t test = the p-value of a t test.
ANSWER: Power of the t test = the p-value of a t test is NOT a true statement. The power is the
probability that a test will reject a false null hypothesis, while the p-value is the exact probability that
a test will (incorrectly) reject a true claim (null hypothesis).
(c) High values of a t test are generally quite important.
(d) Power is the probability that we would reject a false claim (hypothesis).
**5. Matched pairs: 7.43 H0: A = B ( = 0) vs. Ha: A > B ( > 0), where  is the mean difference,
x 0
0.46

 1.581 and the p-value is between 0.10 and 0.05 or 0.074 from Excel.
s / n 0.92 / 10
This is not strong evidence that A is better than B. There is no statistical advantage (higher yield) to
Variety A.
Extra Question: Use your knowledge of x (the sample mean) and s (sample standard deviation) to state
why t tests are not robust against outliers.
ANSWER: Both of these statistics ( x and s) are used to compute a t-value and both use all available
data and are therefore adversely affected by outliers. Both statistics are inflated (or x is deflated if the
outliers are on the left (negative) side) if there are outliers.
(A-B). t 
6. 7.45 The table gives the values for all 50 states, the whole population of states. There is no sampling
involved.
Extra Question: If no sampling is involved and using t test procedures does not make sense, then what
would you do to characterize the mean number of medical doctors per 100,000 people?
ANSWER: It’s been done! We have looked at the entire population, so just calculate the average of all
the data collected. The average of this population is .
Section 7.2
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2
2
s
s
0.7 2 1.82
**7. 7.56 (b) Using df=9, t*=2.262. ( x1  x2 )  t n 1, / 2 1  2 = 4.6  3.2  2.262
=

n1
n2
10
10
(0.019,2.781) . The SE of the difference of the means is 0.61074. From Excel using less conservative
df’s: (0.1837,2.9837) (c) H0: new = old (diff = 0) vs. Ha: new  old (diff  0)
NOTE: Since the 95% confidence interval does not contain 0, we can reject H0 at the 5% level. BUT
JUST BARELY. Since 0 was almost in the interval, if the new monitors cost a lot more, it’s probably
not enough better to offset the cost. However, using Excel’s
& 7.57 A randomized design is generally better. It is possible that new employees will come in batches,
so that we might end up with more employees in one department (with different needs) who get the flat
screens. OR it could be that the next 10 people hate their computers and requested new ones!
**8. 7.58 For the two-bedroom apartments, x = $609 and s=$89.31. For the one-bedroom apartments,
x = $531 and s=$82.79. The SE for the difference is $38.511. Using df=9, t=2.262 and the confidence
interval for the difference is (609  531)  t9,.025
89.312 82.79 2

 ( 9.11,165.11) SPSS output is
10
10
below:
Group Statistics
N
Mean
2
10
1
10
Independent Samples Test
Std.
Std. Error
Deviation
Mean
609.0000 89.3122 28.2430
531.0000 82.7916 26.1810
Levene's Test
for Equality of
Variances
F
Sig.
t-test for Equality of Means
t
df
Sig. Mean Std. Error 95% Confidence Interval
(2- Differen Difference
of the Difference
tailed)
ce
Lower
Upper
.058 78.0000 38.5112
-2.9090
158.9090
Equal
.249
.624 2.025 18
variances
assumed
Equal
2.025 17.898 .058 78.0000 38.5112
-2.9422
158.9422
variances not
assumed
Levene’s Test failed to reject the null hypothesis about equality of variances, therefore it is plausible to
assume that variances are equal and use pooled two-sample t-test.
( n1  1) s1  ( n2  1) s2
9  89.312  9  82.79 2

 86.12 2
n1  n2  2
18
x1  x2
609  531
t

 2.0256
s p 1 / n1  1 / n2 86.12 1 / 10  1 / 10
2
sp 
2
2
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**7.59 (a) H0: 1 = 2 vs. Ha: 1 < 2 , where 1 is the mean cost for one-bedroom apartments and 2 is
the mean cost for two-bedroom apartments. (b) The t-statistic is 2.025 with df=9. The p-value is 0.037.
This is fairly strong evidence that two-bedroom apartments cost more than one-bedroom ones. (c) No,
only that the mean for one-bedrooms is less than the mean for two-bedrooms. (d) Confidence intervals
are generally more useful because they give you a better idea of the size of the difference. The p-value is
a probability and does not tell you the amount of dollars involved.
Not on the assignment:
Extra Question. Also, find the 99% confidence interval for the additional cost of a second bedroom.
What are the key differences between the 95% and 99% intervals?
s12 s22
ANSWER: ( x1  x2  t9,0.005
 ) = ($609 - $531)  3.25 * $38.51 = ($47.16, $203.16). . From
n1 n2
Excel using less conservative df’s: (46.098,202.098) . Key differences between 95% and 99% C.I.’s:
The 99% CI is wider than the 95% CI and includes 0. This 99% interval would not lead us to conclude
that there was a significant change (either increase or decrease) in the subject rents.
9. 7.70 (a) The SE of the difference is 52.47. H0: 1 = 2 vs. Ha: 1 < 2, where 1 is the mean of the
Positive group and 2 is the mean of the Other group. The t-statistic is t 
3118  2733
599 2 672 2

134 5974
 7.34 with
( s1 / n1  s2 / n1 ) 2
 140.61 . The p-value is essentially 0. (b) The 95% interval for
df 
1
1
2
2
( s1 / n1 ) 2 
( s2 / n2 ) 2
n1  1
n2  1
the difference is 282.16 to 487.84. From Excel using less conservative df’s: ( 252.26,517.79) . (c) The
Other group includes women who were not tested. Some of these might be drug users. However, this
would reduce the difference between groups, which is quite significant. On the other hand, there may be
lurking variables. Drug users likely have lower socioeconomic status and generally lower health
condition than non-users. So we cannot conclude from this data that the drug use causes the low birth
weight.
2
2
Section 7.3
10. 7.89 Both numerator and denominator have df=19. The closest table entry is F(20,19). At the 5%
level, the ratio must be at least 2.16.
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