MAT120 Take Home Exam Three
Name
ANSWER KEY!!!
1. You are given the following population: (8, 5, 17).
a) List all the samples with replacement of size 2 given that a selected value can be
replaced and selected a second time. [ 5 pts]
(5, 5) (5, 8) (5, 17)
(8, 8) (8, 5) (8, 17)
(17, 17) (17, 5) (17, 8)
b) Find the sample mean, variance, standard deviation and probability for each of the
samples of size 2 from the population: (8, 5, 17) and then fill in the mean value of each
statistic. [20 pts]
Mean x
Variance s2
Stand. Dev. s
Probability
(5, 5)
5
0
0
.111
(5, 8)
6.5
((-1.5)2 + (1.5)2)/1 =4.5
2.121
.111
(5, 17)
11
((-6)2 + (6)2)/1 = 72
8.485
.111
(8, 8)
8
0
0
.111
(8, 5)
6.5
4.5
2.121
.111
(8, 17)
12.5
6.360
.111
(17, 17)
17
0
(17, 5)
11
72
8.485
.111
(17, 8)
12.5
40.5
6.360
.111
Sample
Mean of
Each Stat.
10.0
((-4.5)2 + (4.5)2)/1= 40.5
26
0
3.77
.111
.111
1
MAT120 Take Home Exam Three
Name
ANSWER KEY!!!
2. You are given that women’s heights are normally distributed with a mean µ of 66.5
inches and a standard deviation of σ of 7.4 inches. [20 pts]
a) If one woman is selected at random find the probability that her height is less than 64
inches?
z = (64 – 66.5) / 7.4 = -2.5 / 7.4 = -.342  .3669
b) If 100 women are randomly selected, find the probability that they have a mean height
of less than 64 inches.
z = (64 – 66.5) / 7.4/10 = -2.5 / .74 = -3.38  .0004
c) If one woman is randomly selected, find the probability that her height is between 67
and 72 inches.
z1 = (67 – 66.5) / 7.4 = .5 / 7.4 = .0676  .53
z2 = (72 – 66.5) / 7.4 = 5.5 / 7.4 = .743  .77
.77 - .53 = .24
d) If 49 women are selected at random, find the probability that they have a mean height
between 67 and 72 inches.
z1 = (67 – 66.5) / 7.4/7 = .5 /1.057 = .473  .6808
z2 = (72 – 66.5) / 7.4 / 7 = 5.5 / 1.057 = 5.20  .9999
.9999 - .6808 = .3191
2
MAT120 Take Home Exam Three
Name
ANSWER KEY!!!
3. A sample of 25 male college basketball players is listed below. Assuming that the
population is normally distributed and the population standard deviation is 7.6 inches,
find a 96% confidence interval for the mean height for all male college basketball
players. [15 pts]
75
68
69
81
68
x = 1883 / 25 = 75.32
69
78
74
86
67
77
84
70
72
81
σx = 7.6 / 5 = 1.52
80
73
75
74
69
82
79
78
76
78
zα/2 = 2.05
CImin = 75.32 – 2.05(1.52) = 75.32 – 3.116 = 72.204
CImax = 75.32 + 3.116 = 78.436
4. What sample size would be needed in number 3 above to reduce the margin of error by
1.5 inches? [10 pts]
3.116 – 1.5 = 1.616
n = (( 2.05) (7.6) / 1.616)2 = (15.580 / 1.616)2 = (9.64)2 = 92.95 = 93
3
MAT120 Take Home Exam Three
Name
ANSWER KEY!!!
5. A sample of 64 gala apples had a mean weight of 145 grams and a sample standard
deviation of 14.2 grams. Find a 98% confidence interval for the mean weight of all gala
apples assuming the weights are normally distributed. [15 pts]
CImin = 145 - tα/2 (s / 8) = 145 – 2.387(14.2/8) = 145 – 4.237 = 140.763
CImax = 145 + 4.237 = 149.237
6. Using a 95% confidence interval, perform the following Hypothesis test.
According to the Beer Institute the mean annual consumption of beer per adult beer
drinker the US is 22.0 gallons. A sample of 25 beer drinkers in New York resulted in a
sample mean of 26.8 gallons. Assume a population standard deviation of 10.5 gallons.
Determine if the mean for New York City differs from the national average. [15 pts]
H0 = µ = μ0 = 22
Ha = µ0 ≠ 22
α = 5 zα/2 = z.025 = 1.96
z = (26.8 – 22) / (10.5/5) = 4.8 / 2.1 = 2.286
Reject the Null Hypothesis in favor of the alternative that NY City differ from the
national average.
4
MAT120 Take Home Exam Three
Name
ANSWER KEY!!!
Bonus Question – 20 points!
Test number 6 above for the hypothesis that beer consumption in New York is greater
than the national average using a sample standard deviation of 12.5 gallons.
H0 = µ = 22.0
Ha = μ > 22.0
α = .05
tα = 1.711
t = (26.8 – 22.0) / 12.5 / 5 = 4.8/2.5 = 1.92
Reject the Null Hypothesis in favor of the alternative that New York consumption is
greater than the national average.
5