A sample of 100 students consists of 60 females and 40 males. The purpose of the study is to determine if the sex of the student is related to passing or failing the first test. Out of the 100 students in the sample, 70 students passed. the expected number of males passing is:
70 |
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40 |
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28 |
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12 |
10 points
QUESTION 10
An engineer who is studying the tensile strength of a steel alloy intended for use in golf club shafts knows that tensile strength is approximately normally distributed with σ = 60 psi. It is believed that the mean of this distribution is 3500 psi. A random sample of 12 specimens has a mean tensile strength of 3450 psi. The correct null for a test of hypothesis on the mean tensile strength would be:
a. |
mu is unequal to 3450 psi |
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b. |
mu equals 3500 psi |
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c. |
x-bar equals 3450 psi |
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d. |
x-bar equals 3500 psi |
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e. |
mu is less than 3500 psi |
10 points
QUESTION 11
The "pooled" estimate of a proportion that results from combining samples will be:
a. |
equal to one of the sample proportions |
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b. |
greater than either sample proportion |
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c. |
between the two sample proportions |
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d. |
less than the smaller of the two sample proportions |
10 points
QUESTION 12
If we want to conduct a test of hypothesis on the difference between two means (large sample), then:
a. |
The sample means should both exceed 30. |
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b. |
Each sample size should exceed 30. |
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c. |
The population mean should not exceed 30. |
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d. |
The sample sizes should not exceed 30. |
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e. |
The combined sample size should exceed 30. |
1) Number of males= 40
Number of females = 60
Total number of students passed= 70
Expected number of males passing = (70/100) * 40 = 0.7 * 40 = 28
2) The correct null hypothesis is
H0:
b. |
mu equals 3500 psi |
3) The "pooled" estimate of a proportion that results from combining samples will be:-
c. |
between the two sample proportions |
4)If we want to conduct a test of hypothesis on the difference between two means (large sample), then:-
b. |
Each sample size should exceed 30. |
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