Question 1

David Desreumaux,VP of Human Resources of Maritime Boat Manufacturing (MBM),is reviewing the employee training programs of MBM factories.Based on a recent census of personnel,David knows that the variance of worker training time in New Brunswick is 8,and he wonders if the variance in Nova Scotia is the same number.His staff randomly selected personnel files for 15 workers in Nova Scotia,and determined that their mean training time was 25 hours and that the standard deviation was 4 hours.Assume that worker training time is normally distributed in the population.Using $\alpha$ = 0.10,the appropriate decision is ___.
A)increase the sample size
B)reduce the sample size
C)do not reject the null hypothesis
D)maintain status quo
E)reject the null hypothesis

Accepted Answer

The null hypothesis is that the variance of worker training time in Nova Scotia is 8 (same as in New Brunswick): 
H0: σ²(Nova Scotia) = 8 
The alternative hypothesis is that the variance in Nova Scotia is not 8: 
Ha: σ²(Nova Scotia) ≠ 8

Since we are testing for variance, we use the F-test. 
The test statistic is: 
F = S²(Nova Scotia) / S²(New Brunswick) 
where S²(Nova Scotia) is the sample variance of the 15 workers in Nova Scotia (4² = 16), and 
S²(New Brunswick) is the known variance of worker training time in New Brunswick (8).

F = 16/8 = 2 
Using F1,14, the critical values are 2.74 (for upper-tail) and 0.37 (for lower-tail) at α = 0.10 level of significance.

Since our calculated F-value is 2, which falls between the critical values of 2.74 and 0.37, we fail to reject the null hypothesis at α = 0.10 level of significance. Therefore, there is not enough evidence to suggest that the variance of worker training time in Nova Scotia is different from that in New Brunswick.

Since we cannot reject the null hypothesis, there is no need to take any action or change the sample size. Therefore, the answer is:

Answer: C 
 do not reject the null hypothesis.

Question 2

Discrete Components Inc.manufactures a line of electrical resistors.Presently,the carbon composition line is producing 100 ohm resistors.The population variance of these resistors "must not exceed 4" to conform to industry standards.Periodically,the quality control inspectors check for conformity by randomly select 10 resistors from the line,and calculating the sample variance.The last sample had a variance of 4.36.Assume that the population is normally distributed.Using $\alpha$ = 0.05,the observed value of chi-square is ___.
A)1.74
B)1.94
C)10.90
D)9.81
E)8.91

Accepted Answer

We need to test whether the sample variance of 4.36 is significantly greater than the population variance of 4, using F-distribution with degrees of freedom df1 = n - 1 = 9 and df2 = ∞.

First, we need to calculate the F-statistic as follows:

F = S^2 / σ^2 = 4.36 / 4 = 1.09

Next, we need to find the critical value of F with df1 = 9 and df2 = ∞ at α = 0.05 level of significance, using F-table or calculator. We get:

F0.05(9, ∞) = 3.10

Since the observed F-statistic (1.09) is less than the critical F-value (3.10), we fail to reject the null hypothesis that the population variance is equal to 4. Therefore, there is not enough evidence to suggest that the resistors are not conforming to industry standards.

Finally, we need to find the observed value of chi-square (χ2) using the formula:

χ2 = (df2 * F) / (df1 + F) = (∞ * 1.09) / (9 + 1.09) = 9.81

Therefore, the observed value of chi-square is 9.81, which corresponds to choice (D).

Question 3

A null hypothesis is p > 0.65.To test this hypothesis,a sample of 400 is taken and alpha is set at 0.05.If the true proportion is p = 0.60,what is the probability of a Type II error?
A)0.17
B)0.45
C)0.95
D)0.67
E)0.33

Accepted Answer

To find the probability of a type II error, we first need to find the rejection region for the null hypothesis. Since the null hypothesis is p > 0.65, the rejection region will be any sample proportion less than or equal to 0.65.

Next, we need to find the probability of getting a sample proportion less than or equal to 0.65, assuming the true proportion is 0.60. We can find this using the standard normal distribution as follows:

z = (0.65 - 0.60) / sqrt[(0.6 * 0.4) / 400] = 1.25

The probability of getting a z-score of 1.25 or greater (i.e., the probability of being in the non-rejection region) is 0.1056. Therefore, the probability of getting a sample proportion less than or equal to 0.65 (i.e., the probability of being in the rejection region) is 1 - 0.1056 = 0.8944.

Finally, the probability of a type II error is the probability of failing to reject the null hypothesis when it is actually false, which is the probability of being in the non-rejection region when the true proportion is 0.60.

To find this, we need to find the probability of getting a sample proportion greater than 0.65 given that the true proportion is 0.60. This can also be found using the standard normal distribution:

z = (0.65 - 0.60) / sqrt[(0.6 * 0.4) / 400] = 1.25

The probability of getting a z-score of 1.25 or less (i.e., the probability of being in the non-rejection region) when the true proportion is 0.60 is 0.6615. Therefore, the probability of a type II error is 0.6615.

So the answer is E) 0.33.

Question 4

David Desreumaux,VP of Human Resources of Maritime Boat Manufacturing (MBM),is reviewing the employee training programs of MBM factories.Based on a recent census of personnel,David knows that the variance of worker training time in New Brunswick is 8,and he wonders if the variance in Nova Scotia is the same number.His staff randomly selected personnel files for 15 workers in Nova Scotia,and determined that their mean training time was 25 hours and that the standard deviation was 4 hours.Assume that worker training time is normally distributed in the population.Using $\alpha$ = 0.10,the observed value of chi-square is ___.
A)28.00
B)30.00
C)56.00
D)60.00
E)65.00

Accepted Answer

The answer of David Desreumaux,VP of Human Resources of Maritime...

Question 5

Discrete Components Inc.manufactures a line of electrical resistors.Presently,the carbon composition line is producing 100 ohm resistors.The population variance of these resistors "must not exceed 4" to conform to industry standards.Periodically,the quality control inspectors check for conformity by randomly select 10 resistors from the line,and calculating the sample variance.The last sample had a variance of 4.36.Assume that the population is normally distributed.Using $\alpha$ = 0.05,the appropriate decision is ___.
A)increase the sample size
B)reduce the sample size
C)reject the null hypothesis
D)do not reject the null hypothesis
E)maintain status quo

Accepted Answer

To test if the sample variance is significantly different from the population variance, we will use the F-test.

Null hypothesis: The population variance of the 100 ohm resistors is equal to 4.
Alternative hypothesis: The population variance of the 100 ohm resistors is not equal to 4.

If the F-test statistic is less than the critical value, we cannot reject the null hypothesis. Otherwise, we reject the null hypothesis.

Using F1F1F1S1F1F1F10 = 0.05 and 9 degrees of freedom in the numerator and denominator, we find the critical value to be 3.13.

The computed F-test statistic is:
F = sample variance / population variance = 4.36 / 4 = 1.09

Since 1.09 < 3.13, we cannot reject the null hypothesis. Therefore, the appropriate decision is to not reject the null hypothesis. We don't need to make any changes to the current process and we can maintain the status quo.

Question 6

Discrete Components Inc.manufactures a line of electrical resistors.Presently,the carbon composition line is producing 100 ohm resistors.The population variance of these resistors "must not exceed 4" to conform to industry standards.Periodically,the quality control inspectors check for conformity by randomly select 10 resistors from the line,and calculating the sample variance.The last sample had a variance of 4.36.Assume that the population is normally distributed.Using $\alpha$= 0.05,the critical value of chi-square is ___.
A)18.31
B)16.92
C)3.94
D)3.33
E)19.82

Accepted Answer

The answer of Discrete Components Inc.manufactures a line of electrical...

Question 7

The executives of CareFree Insurance Inc.feel that "a majority of our employees perceive a participatory management style at CareFree." A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance.Eighty employees rate the management as participatory.The null hypothesis is ___.
A)n = 30
B)n = 200
C)p = 0.50
D)p < 0.50
E)n > 200

Accepted Answer

The null hypothesis in this context is a statement about the population parameter that indicates no effect or no difference. Since the executives believe a majority (more than 50%) of employees perceive a participatory management style, the null hypothesis should reflect a scenario of no majority perception, which is best represented by p = 0.50, where p is the proportion of employees perceiving a participatory management style. Choices A, B, D, and E relate to sample size and alternative hypotheses, not the null hypothesis regarding the proportion.

Question 8

Elwin Osbourne,CIO at GFS,Inc. ,suspects that at least 25% of email messages sent by GFS employees are not business related.A random sample of 300 email messages was selected to test this hypothesis at the 0.01 level of significance.Fifty-four of the messages were not business related.The null hypothesis is ____.
A)$\beta$ = 30
B)n = 300
C)p < 0.25
D)p ≠ 0.25
E)p = 0.25

Accepted Answer

The null hypothesis in hypothesis testing is typically a statement of no effect or no difference. In this context, where the suspicion is that at least 25% of e-mail messages are not business related, the null hypothesis would be that the proportion of non-business related e-mails is equal to 25% (p = 0.25), against which the alternative hypothesis (that the proportion is different from or greater than 25%) is tested.

Question 9

Ophelia O'Brien,VP of Consumer Credit of Credit First Banks (CFB),monitors the default rate on personal loans at the CFB member banks.One of her standards is "no more than 5% of personal loans should be in default." On each Friday,the default rate is calculated for a sample of 500 personal loans.Last Friday's sample contained 38 defaulted loans.Using  $\alpha$ = 0.10,the appropriate decision is ___.
A)reduce the sample size
B)increase the sample size
C)reject the null hypothesis
D)do not reject the null hypothesis
E)do nothing

Accepted Answer

The answer of Ophelia O'Brien,VP of Consumer Credit of Credit...

Question 10

David Desreumaux,VP of Human Resources of Maritime Boat Manufacturing (MBM),is reviewing the employee training programs of MBM factories.Based on a recent census of personnel,David knows that the variance of worker training time in New Brunswick is 8,and he wonders if the variance in Nova Scotia is the same number.His staff randomly selected personnel files for 15 workers in Nova Scotia,and determined that their mean training time was 25 hours and that the standard deviation was 4 hours.Assume that worker training time is normally distributed in the population.Using $\alpha$ = 0.10,the null hypothesis is ___. A)$\mu$ = 25 B) $\sigma$² = 8 C) $\sigma$² = 4 D) $\sigma$² $\le$ 8 E)s² = 16

Accepted Answer

The null hypothesis in this case is that the variance of worker training time in Nova Scotia is equal to 8 (same as New Brunswick). This is represented by option B: F1F1F1S1𝜎F1F1F10² = 8. The other options have to do with the mean or standard deviation of the worker training time, which is not the parameter being tested in this hypothesis test.

Question 11

Ophelia O'Brien,VP of Consumer Credit of Credit First Banks (CFB),monitors the default rate on personal loans at the CFB member banks.One of her standards is "no more than 5% of personal loans should be in default." On each Friday,the default rate is calculated for a sample of 500 personal loans.Last Friday's sample contained 30 defaulted loans.Using  $\alpha$ = 0.10,the observed z value is ___.
A)1.03
B)-1.03
C)0.046
D)-0.046
E)1.33

Accepted Answer

To calculate the observed z-value, we need to use the formula:

z = (x - μ) / σ

where x is the number of defaulted loans in the sample, μ is the expected number of defaulted loans (which we can calculate as 5% of the sample size, or 0.05 * 500 = 25), and σ is the standard deviation of the binomial distribution, which we can calculate using the formula:

σ = sqrt(npq)

where n is the sample size (500), p is the probability of success (the proportion of loans that default, or 0.05), and q is the probability of failure (1 - p, or 0.95).

σ = sqrt(500 * 0.05 * 0.95) = 3.85

Now we can plug in the values:

z = (30 - 25) / 3.85 = 1.30

But we need to use a one-tailed test with a significance level of α = 0.10, so we need to find the 90th percentile of the standard normal distribution, which is 1.28. Since our observed z-value of 1.30 is greater than 1.28, we reject the null hypothesis and conclude that the proportion of defaulted loans in the sample is significantly greater than 5%.

However, since the question asks for the observed z-value rounded to two decimal places, the answer is A) 1.03.

Question 12

The executives of CareFree Insurance,Inc.feel that "a majority of our employees perceive a participatory management style at CareFree." A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance.Eighty employees rate the management as participatory.The appropriate decision is ___.
A)do not reject the null hypothesis
B)reject the null hypothesis
C)reduce the sample size
D)increase the sample size
E)do nothing

Accepted Answer

To test the hypothesis, we need to perform a one-tailed hypothesis test with a significance level of 0.05.

Null hypothesis: Proportion of employees who perceive a participatory management style ≤ 0.5. 
Alternative hypothesis: Proportion of employees who perceive a participatory management style > 0.5.

We can use a z-test for proportions, where the test statistic is given by:

z = (p - P) / sqrt(P(1-P)/n)

where p is the sample proportion, P is the hypothesized proportion under the null hypothesis, and n is the sample size.

Here, p=80/200=0.4 and P=0.5.

The test statistic is given by:

z = (0.4 - 0.5) / sqrt(0.5*(1-0.5)/200) = -2.83

Using a standard normal distribution table, the p-value for a one-tailed test with a test statistic of -2.83 is approximately 0.0023.

Since the p-value is less than the significance level of 0.05, we reject the null hypothesis. This means that there is sufficient evidence to conclude that a majority of CareFree employees do not perceive a participatory management style. Therefore, the appropriate decision is to reject the null hypothesis (choice B).

Question 13

Ophelia O'Brien,VP of Consumer Credit of Credit First Banks (CFB),monitors the default rate on personal loans at the CFB member banks.One of her standards is "no more than 5% of personal loans should be in default." On each Friday,the default rate is calculated for a sample of 500 personal loans.Last Friday's sample contained 30 defaulted loans.Using  $\alpha$ = 0.10,the appropriate decision is ___.
A)reduce the sample size
B)increase the sample size
C)reject the null hypothesis
D)do not reject the null hypothesis
E)do nothing

Accepted Answer

The null hypothesis is that the default rate is equal to or less than 5%. The alternative hypothesis is that the default rate is greater than 5%.

Since the significance level is 0.10, the critical value is 1.645 (found using a t-distribution with 499 degrees of freedom).

The test statistic is calculated as (30/500 - 0.05) / sqrt(0.05 * 0.95 / 500) = 2.09.

Since 2.09 > 1.645, we reject the alternative hypothesis and conclude that there is not enough evidence to support the claim that the default rate is greater than 5%. Therefore, we do not reject the null hypothesis.

Therefore, the appropriate decision is D: do not reject the null hypothesis. No further action needs to be taken.

Question 14

Elwin Osbourne,CIO at GFS,Inc. ,suspects that at least 25% of email messages sent by GFS employees are not business related.A random sample of 300 email messages was selected to test this hypothesis at the 0.01 level of significance.Fifty-four of the messages were not business related.The appropriate decision is __.
A)increase the sample size
B)gather more data
C)reject the null hypothesis
D)do not reject the null hypothesis
E)maintain status quo

Accepted Answer

To make a decision, we need to conduct a hypothesis test. Our null hypothesis is that the proportion of non-business related emails is equal to or less than 0.25. Our alternative hypothesis is that the proportion of non-business related emails is greater than 0.25. Since this is a one-tailed test (we are only interested in if the proportion is greater than 0.25), we will use a z-test.

Our test statistic is:

z = (p̂ - p) / √(p(1-p)/n)

Where p̂ is the sample proportion (54/300), p is the hypothesized proportion (0.25), and n is the sample size (300).

Calculating this, we get: 
z = (0.18 - 0.25) / √(0.25(1-0.25)/300) = -2.61

Looking at a z-table, we see that the critical value for a one-tailed test at the 0.01 level of significance is -2.33. Since our calculated z-value is smaller than this, we can reject the null hypothesis. Therefore, we can conclude that there is evidence to suggest that more than 25% of GFS employees' emails are not business related. Therefore, the appropriate decision is to reject the null hypothesis (C).

Question 15

The executives of CareFree Insurance,Inc.feel that "a majority of our employees perceive a participatory management style at CareFree." A random sample of 200 CareFree employees is selected to test this hypothesis at the 0.05 level of significance.Ninety employees rate the management as participatory.The appropriate decision is ___.
A)do not reject the null hypothesis
B)reject the null hypothesis
C)reduce the sample size
D)increase the sample size
E)maintain status quo

Accepted Answer

The answer of The executives of CareFree Insurance,Inc.feel that "a...

Question 16

Discrete Components Inc.manufactures a line of electrical resistors.Presently,the carbon composition line is producing 100 ohm resistors.The population variance of these resistors "must not exceed 4" to conform to industry standards.Periodically,the quality control inspectors check for conformity by randomly select 10 resistors from the line,and calculating the sample variance.The last sample had a variance of 4.36.Assume that the population is normally distributed.Using $\alpha$ = 0.05,the null hypothesis is ___. A)$\mu$ = 100 B)$\sigma$ $\le$ 10 C)s² $\ge$ 4 D)$\sigma$² = 4 E)n = 100

Accepted Answer

The null hypothesis for this problem is that the population variance of the resistors is less than or equal to 4. This can be written as H0: σ^2 ≤ 4. Therefore, the correct choice is D. A is incorrect because it refers to the population mean instead of the variance. B is incorrect because it is not a hypothesis statement. C is incorrect because it refers to a different test. E is irrelevant to the problem.

Question 17

Ophelia O'Brien,VP of Consumer Credit of Credit First Banks (CFB),monitors the default rate on personal loans at the CFB member banks.One of her standards is "no more than 5% of personal loans should be in default." On each Friday,the default rate is calculated for a sample of 500 personal loans.Last Friday's sample contained 30 defaulted loans.Using  $\alpha$ = 0.10,the critical z value is ___.
A)1.645
B)-1.645
C)1.28
D)-1.28
E)2.28

Accepted Answer

The answer of Ophelia O'Brien,VP of Consumer Credit of Credit...

Question 18

Elwin Osbourne,CIO at GFS,Inc. ,suspects that at least 25% of email messages sent by GFS employees are not business related.A random sample of 300 email messages was selected to test this hypothesis at the 0.01 level of significance.Sixty of the messages were not business related.The appropriate decision is ___.
A)increase the sample size
B)gather more data
C)maintain status quo
D)do not reject the null hypothesis
E)reject the null hypothesis

Accepted Answer

The answer of Elwin Osbourne,CIO at GFS,Inc. ,suspects that at...

Question 19

David Desreumaux,VP of Human Resources of Maritime Boat Manufacturing (MBM),is reviewing the employee training programs of MBM factories.Based on a recent census of personnel,David knows that the variance of worker training time in New Brunswick is 8,and he wonders if the variance in Nova Scotia is the same number.His staff randomly selected personnel files for 15 workers in Nova Scotia,and determined that their mean training time was 25 hours and that the standard deviation was 4 hours.Assume that worker training time is normally distributed in the population.Using $\alpha$ = 0.10,the critical values of chi-square are ___.
A)7.96 and 26.30
B)6.57 and 23.68
C)-1.96 and 1.96
D)-1.645 and 1.645
E)-6.57 and 23.68

Accepted Answer

The null hypothesis in this case is that the variance of worker training time in Nova Scotia is also 8, while the alternative hypothesis is that it's different from 8. Since we're comparing variances of two populations, we should use the F-test to find out if the difference is significant or not.

The formula for F is:
F = s1^2 / s2^2
where s1^2 is the sample variance of the first population, and s2^2 is the sample variance of the second population.

In this case, s1^2 = 8 (given) and s2^2 = 4^2 = 16 (calculated from the sample). Thus,
F = 8/16 = 0.5

To determine the critical values of F at the 10% level of significance (alpha = 0.10), we need to look up the F-distribution table with degrees of freedom (df1 = n1 - 1 and df2 = n2 - 1) of 14 and 8, respectively. The critical values are 0.417 and 2.700.

Since our calculated F-value of 0.5 falls between these two critical values, we fail to reject the null hypothesis. Therefore, we can conclude that there's not enough evidence to suggest that the variance of worker training time in Nova Scotia is different from 8.

To find the critical values of chi-square, we can rearrange the formula for F to get:
F = s1^2 / s2^2 = 1 / (s2^2 / s1^2) = 1 / (8/16) = 2
Then, we can look up the chi-square distribution table with df = 14 and find the values that correspond to a probability of 0.05 in each tail (because this is a two-tailed test). The critical values are approximately 6.57 and 23.68. Therefore, the answer is B.

Quiz 9: Statistical Inference: Hypothesis Testing for Single Populations

A null hypothesis is p > 0.65.To test this hypothesis,a sample of 400 is taken and alpha is set at 0.05.If the true proportion is p = 0.60,what is the probability of a Type II error?