10.20. Some studies have shown that in the United States, men spend more than wo
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Question
10.20. Some studies have shown that in the United States, men spend more than women buying
gifts and cards on Valentine's Day. Suppose a researcher wants to test this hypothesis by
randomly sampling nine men and 10 women with comparable demographic
characteristics from various large cities across the United States to be in a study. Each
study participant is asked to keep a log beginning one month before Valentine's Day and
record all purchases made for Valentine's Day during that one-month period. The
resulting data are shown below. Use these data and a 1% level of significance to test to
determine if, on average, men actually do spend significantly more than women on
Valentine's Day. Assume that such spending is normally distributed in the population and
that the population variances are equal.
Men Women
$107.48 $125.98
143.61 45.53
90.19 56.35
125.53 80.62
70.79 46.37
83.00 44.34
129.63 75.21
154.22 68.48
93.80 85.84
126.11
a) Decision is:
b) Test t =
c) p-value for this test is =
d) Critical t value(s) is/are
e) Copy and paste a little distribution diagram and label the critical t value(s)
f) Copy
Also repeat the calculation using the Excel routine for NON-EQUAL population variances to
show any differences from above.
a) Decision is:
b) Test t =
c) p-value for this test is =
d) Critical t value(s) is/are
Note the equal-variances case is accepted as sufficiently accurate for nearly all t-tests mentioned in
Chapter 10. The equal-variances case always has an integer and easy-to-calculate degrees of freedom.
Explanation / Answer
For men, x-bar = 110.917 and s = 28.792
For women, x-bar = 75.463 and s = 30.530
n1 = 9, n2 = 10, x1-bar = 110.916667, x2-bar = 75.463, s1 = 28.7915925, s2 = 30.5296526
Ha: 1 2
Ha: 1 > 2
Degrees of freedom = 9 + 10 - 2 = 17
Critical t- score = 1.73960672
We will reject Ho if t > 1.73960672
Test Statistic:
Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = (((9 - 1) * 28.7915925228182^2 + (10 - 1) * 30.5296526347746^2)/(9 + 10 - 2)) = 29.72440432
SE = s * {(1 /n1) + (1 /n2)} = 29.7244043229287 * ((1/9) + (1/10)) = 13.65742128
t = (x1-bar -x2-bar)/SE = (110.916666666667 - 75.463)/13.6574212771363 = 2.595926855
p- value = 0.00941903
Decision:
Since 2.59592686 > 1.739606716 we reject Ho and accept Ha
Conclusion:
There is sufficient evidence that men actually do spend significantly more than women on Valentine’s Day.
Data Analysis Output:
t-Test: Two-Sample Assuming Equal Variances
Men
Women
Mean
110.9166667
75.463
Variance
828.9558
932.05969
Observations
9
10
Pooled Variance
883.5402124
Hypothesized Mean Difference
0
df
17
t Stat
2.595926855
P(T<=t) one-tail
0.009419029
t Critical one-tail
1.739606716
P(T<=t) two-tail
0.018838057
t Critical two-tail
2.109815559
For men, x-bar = 110.917 and s = 28.792
For women, x-bar = 75.463 and s = 30.530
n1 = 9, n2 = 10, x1-bar = 110.916667, x2-bar = 75.463, s1 = 28.7915925, s2 = 30.5296526
Ha: 1 2
Ha: 1 > 2
Degrees of freedom = 9 + 10 - 2 = 17
Critical t- score = 1.73960672
We will reject Ho if t > 1.73960672
Test Statistic:
Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = (((9 - 1) * 28.7915925228182^2 + (10 - 1) * 30.5296526347746^2)/(9 + 10 - 2)) = 29.72440432
SE = s * {(1 /n1) + (1 /n2)} = 29.7244043229287 * ((1/9) + (1/10)) = 13.65742128
t = (x1-bar -x2-bar)/SE = (110.916666666667 - 75.463)/13.6574212771363 = 2.595926855
p- value = 0.00941903
Decision:
Since 2.59592686 > 1.739606716 we reject Ho and accept Ha
Conclusion:
There is sufficient evidence that men actually do spend significantly more than women on Valentine’s Day.
Data Analysis Output:
t-Test: Two-Sample Assuming Equal Variances
Men
Women
Mean
110.9166667
75.463
Variance
828.9558
932.05969
Observations
9
10
Pooled Variance
883.5402124
Hypothesized Mean Difference
0
df
17
t Stat
2.595926855
P(T<=t) one-tail
0.009419029
t Critical one-tail
1.739606716
P(T<=t) two-tail
0.018838057
t Critical two-tail
2.109815559
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