Question 1

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the sample proportion?
A) 0.41
B) 0.36%
C) 0.41%
D) 0.36

Accepted Answer

The sample proportion is the percentage of viewers in the sample who watch the CBS affiliate's 11 p.m. news, which is given as 36%. Therefore, the answer is D, which is 0.36. Note that choices A and C refer to the percentage of the overall viewing audience reached by the newscast, not the proportion of viewers in a specific sample. Choice B is incorrect because it mistakenly expresses 36% as a decimal.

Question 2

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. If α = 0.025, what will be the decision?
A) Accept the null.
B) Reject the null and conclude the defects are not greater than 10%.
C) Reject the null and conclude the defects are greater than 10%.
D) Fail to reject the null.

Accepted Answer

The answer of It is claimed that in a bushel...

Question 3

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is your decision if α = 0.01?
A) Fail to reject the null hypothesis.
B) Reject the null hypothesis and conclude the newscast does not reach 41% of the audience.
C) Fail to reject the alternate and conclude the newscast does not reach 41% of the audience.
D) Reject the alternate and conclude the newscast reaches about 41% of the audience.

Accepted Answer

We have a null hypothesis that the proportion of viewers who watch the late evening news on the local CBS affiliate is equal to 0.41. Our alternative hypothesis is that it is less than 0.41.

H0: p = 0.41
Ha: p < 0.41

We are given a sample of 100 viewers, of which 36 watch the late evening news on the local CBS affiliate. We can use a one-sample proportion test to compare the sample proportion to the hypothesized proportion.

The test statistic is:

z = (p̂ - p) / sqrt(p(1-p) / n)
  = (0.36 - 0.41) / sqrt(0.41(1-0.41) / 100)
  = -1.33

Using a one-tailed test at the 0.01 level of significance, the critical value is -2.33 (based on a standard normal distribution). Since -1.33 is greater than -2.33, we fail to reject the null hypothesis. In other words, we do not have enough evidence to conclude that the proportion of viewers who watch the late evening news on the local CBS affiliate is less than 0.41.

Therefore, the best choice is A) Fail to reject the null hypothesis.

Question 4

Students were sampled in order to determine their support for the legalization of gambling in their community. A sample of 150 students were asked whether or not they supported legalization of gambling, and the following results were obtained.  The value of the chi-square test statistic equals _____.
A) -3
B) -2
C) +3
D) 4

Accepted Answer

We cannot calculate the chi-square test statistic based on the given information alone. We need to know the number of students who supported and opposed the legalization of gambling in order to construct a contingency table and calculate the expected frequencies. Once we have the observed and expected frequencies, we can use the formula for chi-square test statistic:

χ2 = Σ (O - E)2 / E

where O = observed frequency, E = expected frequency, and Σ = the sum across all cells of the contingency table.

Without this information, we cannot determine the value of the chi-square test statistic. Therefore, none of the answer choices is correct.

Question 5

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the critical value if α = 0.01?
A) +2.576
B) +2.326
C) ±2.576
D) -2.326

Accepted Answer

To determine the critical value we need to perform a hypothesis test for the proportion.

Null hypothesis: p=0.41
Alternative hypothesis: p<0.41 (since we are looking for the critical value at alpha=0.01)

We can use the normal distribution to find the critical value, since the sample size (100) is large enough to satisfy the conditions of the central limit theorem.

Using a standard normal distribution table, the critical value for a left-tailed test at alpha=0.01 is approximately -2.576.

Therefore, the answer is C (±2.576), since we are looking for the critical value in the left-tail of the distribution.

Question 6

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the critical value if the level of significance is 0.10?
A) -1.282
B) ±1.645
C) -2.576
D) +2.576

Accepted Answer

To determine the critical value, we need to use a one-tailed test because the claim is that the proportion of viewers is greater than 41%. The level of significance is 0.10, so the area in the right-tail is 0.10. Using a standard normal distribution table, the critical value for a right-tailed test with α = 0.10 is ±1.645. Therefore, the best choice is B.

Question 7

In the past five years, 45% of the tourists who visited Orlando, Florida, went to see local attractions. The city council recently spent a significant amount of money on advertising and promoting visits to area attractions. They are interested in knowing whether the advertising campaign was effective-whether it increased the proportion of tourists visiting local attractions. The proper set of hypotheses is ________________. A) H₀: p > 0.45 H₁: p ≤ 0.45 B) H₀: p < 0.45 H₁: p ≥ 0.45 C) H₀: p ≥ 0.45 H₁: p < 0.45 D) H₀: p ≤ 0.45 H₁: p > 0.45

Accepted Answer

The null hypothesis should always be the claim that is being tested, and in this case, it is the proportion of tourists who visit local attractions. Since the city council is interested in knowing whether the advertising campaign increased the proportion of tourists visiting local attractions, we can write: H₀: p ≤ 0.45 (the null hypothesis that the proportion is less than or equal to 0.45) H₁: p > 0.45 (the alternative hypothesis that the proportion is greater than 0.45) Therefore, the correct set of hypotheses is D, H₀: p ≤ 0.45 H₁: p > 0.45, which states that the proportion of tourists visiting local attractions is not greater than or equal to 0.45, and the alternative hypothesis is that the proportion is greater than 0.45, indicating an increase in the proportion of tourists after the advertising campaign.

Question 8

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the z-test statistic?
A) +0.125
B) +0.278
C) -1.645
D) +1.667

Accepted Answer

The z-test statistic is calculated using the formula: z = (p̂ - p) / √(p(1-p)/n), where p̂ = 50/400 = 0.125, p = 0.10, and n = 400. Plugging in these values gives z = (0.125 - 0.10) / √(0.10*0.90/400) = +1.667.

Question 9

A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. Thus, the alternate hypothesis is ___________________.

Accepted Answer

The answer of A manufacturer claims that less than 1%...

Question 10

The degrees of freedom for a contingency table with 6 rows and 3 columns is ______.
A) 18
B) 9
C) 10
D) 3

Accepted Answer

The formula to calculate degrees of freedom for a contingency table is:
(df) = (number of rows - 1) x (number of columns - 1)

Plugging in the values given in the question:
(df) = (6 - 1) x (3 - 1) 
(df) = 5 x 2 
(df) = 10

Therefore, the degrees of freedom for the contingency table is 10. The best choice is C.

Question 11

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the sample proportion?
A) 0.10
B) 0.125
C) 40
D) 0.40

Accepted Answer

The sample proportion is calculated by dividing the number of defective peaches by the total number of peaches in the sample: 50/400 = 0.125.

Question 12

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the null hypothesis? A) H₀: π ≠ 0.10 B) H₀: π ≥ 0.10 C) H₀: π ≤ 0.10 D) H₀: π < 0.10

Accepted Answer

The null hypothesis should always state that there is no significant difference between the sample proportion and the hypothesized proportion. In this case, the hypothesized proportion is less than 10% defective peaches in a bushel, which means the null hypothesis should state that the true proportion is at least 10% (or greater). Therefore, the correct null hypothesis is H₀: π ≥ 0.10, which is represented by choice B.

Question 13

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the alternate hypothesis? A) H₁: π = 0.36 B) H₁: π = 0.41 C) H₁: π ≠ 0.41 D) H₁: µ ≠ 0.41

Accepted Answer

The alternate hypothesis for this situation would be H1: the true proportion of viewers who watch the late evening news on the local CBS station is not equal to 41%. This is represented as H₁: π ≠ 0.41, making choice C the correct answer.

Question 14

Some important uses of the chi-square distribution include:
A) testing for goodness-of-fit.
B) determining whether a variable appears to follow some specified distribution.
C) testing for the association of two categorical variables.
D) All of these are true.

Accepted Answer

The answer of Some important uses of the chi-square distribution...

Question 15

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the z test statistic?
A) 1.02
B) 1.22
C) -1.02
D) -1.22

Accepted Answer

The z-test statistic is calculated using the formula $z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$, where $\hat{p}$ is the sample proportion, $p$ is the population proportion, and $n$ is the sample size. Here, $\hat{p} = 0.36$, $p = 0.41$, and $n = 100$. Plugging these values into the formula gives $z = \frac{0.36 - 0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}} = -1.02$.

Question 16

Students were sampled in order to determine their support for the legalization of gambling in their community. A sample of 150 students were asked whether or not they supported legalization of gambling, and the following results were obtained.  The number of degrees of freedom associated with this scenario is _____.
A) 149
B) 2
C) 150
D) 3

Accepted Answer

The answer of Students were sampled in order to determine...

Question 17

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the alternate hypothesis for a one-sided test? A) H₁: π ≥ 0.10 B) H₁: π > 0.10 C) H₁: π ≤ 0.10 D) H₁: π < 0.10

Accepted Answer

The answer of It is claimed that in a bushel...

Question 18

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the critical value for α = 0.025?
A) +1.960
B) ±1.645
C) -1.960
D) -1.645

Accepted Answer

To find the critical value for α = 0.025, we need to use a Z-table. Since the sample size is greater than 30, we can use the normal distribution. 
First, we need to find the sample proportion of defective peaches: 
p̂ = 50/400 = 0.125
Next, we can calculate the standard error of the sample proportion: 
SE = sqrt(p̂(1-p̂)/n) = sqrt(0.125(1-0.125)/400) = 0.0221
Using the Z-table or calculator, we can find the critical value for α = 0.025 to be -1.960. 
Therefore, the answer is C.

Question 19

A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. Thus, the null hypothesis is ___________________.

Accepted Answer

The answer of A manufacturer claims that less than 1%...

Question 20

For a contingency table test of independence, the formula to compute the degrees of freedom is __________________________________.

Accepted Answer

The answer of For a contingency table test of independence,...

Quiz 15: Nonparametric Methods: Nominal Level Hypothesis Tests

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the sample proportion?

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. If α = 0.025, what will be the decision?

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is your decision if α = 0.01?

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the critical value if α = 0.01?

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the z-test statistic?

A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. Thus, the alternate hypothesis is ___________________.

The degrees of freedom for a contingency table with 6 rows and 3 columns is ______.

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the sample proportion?

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the null hypothesis?

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the alternate hypothesis?

Some important uses of the chi-square distribution include:

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the z test statistic?

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the alternate hypothesis for a one-sided test?

It is claimed that in a bushel of peaches, less than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the critical value for α = 0.025?

A manufacturer claims that less than 1% of all its products do not meet the minimum government standards. A survey of 500 products revealed that 10 did not meet the standard. Thus, the null hypothesis is ___________________.

For a contingency table test of independence, the formula to compute the degrees of freedom is __________________________________.