There is evidence that men actually do spend significantly
more than women on Valentine’s Day.
Work
Show all your work for the questions below.
Question 1
μ = 75, σ = 10, x = 85
z = (x – μ)/σ
= (85 – 75)/10
= 1
Question 2
μ = 73, σ = 11
(a) x = 95

z = (x – μ)/σ
= (95 – 73)/11
= 2
P(x < 95) = P(z < 2) = 0.9772
(b) x1 = 62 and x2 = 84
z1 = (62 – 73)/11 = -1 and z2 = (84 – 73)/11 = 1
P(62 < x < 84) = P(-1 < z < 1) = 0.6827
Question 3
π = 0.19,n = 950
(a) p = 0.20
z = (p – π)/{√π * (1 – π)/n}
= (0.20 – 0.19)/{√0.19* (1 – 0.19)/950}
= 0.7857
P(p > 0.20) = P(z > 0.7857) = 0.2160
(b) p1 = 0.18 and p2 = 0.21
z1 = (0.18 – 0.19)/{√0.19* (1 – 0.19)/950} = -0.7857 and
z2 = (0.21 – 0.19)/{√0.19 * (1 – 0.19)/950} = 1.5713
P(0.18 < π < 0.21) = P(-0.7857 < z < 1.5713) = 0.7259
Question 4
3.27 3.29 3.16 3.20 3.37
3.20 3.23 3.19 3.20 3.24
3.16 3.07 3.27 3.09 3.35
3.15 3.23 3.14 3.05 3.35
3.21 3.14 3.14 3.07 3.10
For the given data, x-bar = 3.173 and s = 0.069
n =
25

μ =
3.16
s =
0.069
x-bar =
3.173
Hypotheses:
Ho: μ ≤
3.16
Ha: μ >
3.16
Decision Rule:
α =
0.01
Degrees of freedom =
25 - 1 =
24
Critical t- score =
2.492159469
Reject Ho if t > 2.492159469
Test Statistic:
SE = s/n =
0.069/√25 =
0.0138
t = (x-bar - μ)/SE =
(3.173 - 3.16)/0.0138 =
0.942028986
p- value =
0.177782159
Decision (in terms of the hypotheses):
Since
0.942028986 < 2.492159469 we fail to reject Ho
Conclusion (in terms of the problem):
There is no sufficient evidence to reject the claim. The mean price may not be greater than $3.16.
Question 5
n =
67
p =
0.47
p' =
40/67 = 0.597
Hypotheses:
Ho: p =
0.47
Ha: p ≠
0.47

Decision Rule:
α =
0.05
Lower Critical z- score =
-1.9600
Upper Critical z- score = 1.9600
Reject Ho if |z| >
1.9600
Test Statistic:
SE = p (1 - p)/n =
√(0.47 * (1 - 0.47)/67) =
0.0610
z = (p'- p)/SE = (0.597014925373134 - 0.47)/0.0609746705424572 = 2.0831
p- value = 0.0372
Decision (in terms of the hypotheses):
Since 2.0831 > 1.9600 we reject Ho and accept Ha
Conclusion (in terms of the problem):
There is t evidence to reject the findings of Robert Half International. It appears that the proportion of
CFOs who get their money news from the newspapers is not 47%.
Question 6
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
For the 9 men, x-bar = 110.917 and s = 28.792
For the 10 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
Hypotheses:
Ha: μ1 ≤ μ2
Ha: μ1 > μ2
Decision Rule:
α = 0.05
Degrees of freedom = 9 + 10 - 2 = 17
Critical t- score = 1.73960672
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

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- Fall '19
- Null hypothesis, Statistical hypothesis testing, reject null