Exercise 2 - PM 299

PM299.
Exercise 2
Date Due: 25 October 2019

1. 20 points
The following table shows the grouped data, in classes, for the heights of 50 people.
height (in cm) - classes frequency
120 <- 130 2
130 <- 140 5
140 <- 150 25
150 <- 160 10
160 <- 170 8
a) Calculate the mean of the height of the 50 people.

height m f mf
120 ≤130 125 2 250
130 ≤ 140 135 5 675
140 ≤ 150 145 25 3625
150 ≤ 160 155 10 1550
160 ≤ 170 165 8 1320
∑ mf = 7420
Mean = μ = ∑ mf
Where: m = midpoint of the class
f = frequency
N = number of cases
μ= 7420 = 148.4 cm
50
b) Calculate the standard deviation of the height of 50 people.
x x-x (x-x)2 f f (x-x) f (x-x)2
125 -23.4 547.56 2 -46.8 1095.12
135 -13.4 179.56 5 -67 897.8
145 -3.4 11.56 25 -85 289
155 6.6 43.56 10 66 435.6
165 16.6 275.56 8 132.8 _ 2204.48
∑ f = 50 ∑ f (x-x) = 0 ∑ f (x-x)2 = 4922
Standard deviation =
σ = √𝛴𝑓𝑥 − 𝑥)2
𝛴𝑓
c) Interpret your result.
There is no variation in the heights of 50 people. The heights of the 50 people do not vary.
Fill-in the Blanks

2. 15 points
A researcher knows that the average height of Filipino women is 1.525 meters. A
random sample of 26 women was taken and was found to have a mean height of
1.56 meters with a standard deviation of .10 meters.
Answer the following:
1. What test statistic should be used? One sample T- test
2. Formulate the null hypothesis. H0; μ = 1.525

3. Formulate the alternative hypothesis Ha: μ ≠ 1.525
4. At .01 level of significance, what is table value? 2.4049
5. At .05 level of significance, what is the table value? 1.6766
3. 15
Eggs laid by chicken are known to have lengths normally distributed, with a mean of
6 cm and standard deviation of 1.4 cm. What is the probability of finding an egg
bigger than 8 cm. in length? ____________
μ = 6 cm
σ = 1.4
μ> 8 cm
Z = 0.923 or 92.3%
4. 20
The lifetime of a tire brand is approximately normally distributed with mean 42,500
miles and standard deviation 3,200 miles. What percentage of tires last over 45,000
miles? Show your computation.
a) μ = 42,500
σ = 3,200
x = 45,000
Z=x=u
= 45,000 – 42,500
3200
= 0.78125 ≈ 78.13%
The blood cholesterol levels in 20-year old students is normally distributed with a mean
of 165 milligrams of cholesterol per deciliter of blood and a standard deviation of 30.
What percentage of 20-year old students have level between 120 and 200? Show your
computation.
μ = 165
σ= 30
120 < x < 200
@ x = 120; Z = 120 - 165 = -1.5
30
@ x = 200; Z = 200 - 165 = 1.167
30
P (120 < x < 200) = P (-1.5 < Z < 1.167)
0.87900 – 0.06681 = 0.81219 ≈ 81.22%
5. 30 points
A training was given to 30 local entrepreneurs to increase their monthly sales. Their
mean monthly sale is 200 units prior to the training. Six months after the completion of
the training, a survey of the trainees revealed a mean-monthly sale of 215 units with a
standard deviation of 15 units. Answer the questions below: (NOTE: Include
computation where needed.)
Do a hypothesis test to determine whether the sales of the trainees increased. Use α =
0.05.
a. What is the null hypothesis?

A. Ho: µ> 215
B. Ho: µ≠ 215 answer
C. Ho: µ< 200
D. Ho: µ≤ 200
E. Ho: µ> 200
b. What is the alternative hypothesis?
A. Ho: µ> 215
B. Ho: µ≠ 215
C. Ho: µ≤ 200 answer
D. Ho: µ< 200
E. Ho: µ> 200
c. If you are going to use t-test, what is the direction of the test?
A. Right tailed answer

B. Left tailed
C. Two-tailed
D. Nine-tailed
E. None of the above
d. Given the critical values below, what is the right decision

rule? One-tail critical value at α = 0.05 is 1.699
Two-tail critical value at α = 0.05 is 2.045
A. Reject Ho if tcomputed > 2.045, otherwise fail to reject Ho.

B. Reject Ho if tcomputed > 1.699, otherwise fail to reject Ho. answer
C. Reject Ho if tcomputed < -2.045, otherwise fail to reject Ho.

D. Reject Ho if tcomputed < -1.699, otherwise fail to reject Ho.
E. None of the above
e. What should be your decision?

A. Reject Ho
B. Fail to reject Ho answer

C. Reject Ha
D. Fail to reject Ha
f. Based on your decision in no. 5, what is the conclusion of your test.
There is no significant difference between the mean monthly sales prior and after
training.

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PM299.

Date Due: 25 October 2019

height (in cm) - classes frequency

120 <- 130 2

130 <- 140 5

140 <- 150 25

150 <- 160 10

160 <- 170 8

a) Calculate the mean of the height of the 50 people.

Where: m = midpoint of the class

b) Calculate the standard deviation of the height of 50 people.

x x-x (x-x)2 f f (x-x) f (x-x)2

125 -23.4 547.56 2 -46.8 1095.12

135 -13.4 179.56 5 -67 897.8

145 -3.4 11.56 25 -85 289

155 6.6 43.56 10 66 435.6

165 16.6 275.56 8 132.8 _ 2204.48

∑ f = 50 ∑ f (x-x) = 0 ∑ f (x-x)2 = 4922

c) Interpret your result.

Fill-in the Blanks

Answer the following:

1. What test statistic should be used? One sample T- test

2. Formulate the null hypothesis. H0; μ = 1.525

120 < x < 200

@ x = 120; Z = 120 - 165 = -1.5

@ x = 200; Z = 200 - 165 = 1.167

P (120 < x < 200) = P (-1.5 < Z < 1.167)

0.87900 – 0.06681 = 0.81219 ≈ 81.22%

a. What is the null hypothesis?

B. Ho: µ≠ 215 answer

C. Ho: µ< 200

E. Ho: µ> 200

b. What is the alternative hypothesis?

A. Ho: µ> 215

C. Ho: µ≤ 200 answer

D. Ho: µ< 200

E. Ho: µ> 200

A. Right tailed answer

d. Given the critical values below, what is the right decision

Two-tail critical value at α = 0.05 is 2.045

A. Reject Ho if tcomputed > 2.045, otherwise fail to reject Ho.

C. Reject Ho if tcomputed < -2.045, otherwise fail to reject Ho.

e. What should be your decision?

B. Fail to reject Ho answer

f. Based on your decision in no. 5, what is the conclusion of your test.

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