Question 1

If 20 out of 50 students sampled live in a college dormitory, what is the estimated proportion of students at the University living in a dormitory?
A) 0.20
B) 0.40
C) 0.50
D) 0.60

Accepted Answer

The estimated proportion of students at the University living in a dormitory can be found by taking the number of students who live in a dormitory (20) divided by the total number of students sampled (50): 
20/50 = 0.4 
Therefore, the answer is B) 0.40.

Question 2

If the decision is to reject the null hypothesis of no difference between two population proportions at the 5% level of significance, what are the alternative hypothesis and rejection region? A) H₁: π₁ ≠ π₂; z > +1.645 and z < -1.645 B) H₁: π₁ ≠ π₂; z > +1.960 and z < -1.960 C) H₁: π₁ > π₂; z < -1.645 D) H₁: π₁ > π₂; z < -1.960

Accepted Answer

The alternative hypothesis is that the two population proportions are not equal (π1 ≠ π2). The rejection region for a 5% level of significance using a two-tailed test is z > +1.960 or z < -1.960. Therefore, option B is the best choice as it includes the correct alternative hypothesis and rejection region.

Question 3

The claim that "40% of those persons who retired from an industrial job before the age of 60 would return to work if a suitable job was available" is to be investigated at the 0.02 significance level. If 74 out of the 200 workers sampled said they would return to work, what is our decision?
A) Do not reject the null hypothesis because -0.879 lies in the region between +2.326 and -2.326.
B) Do not reject the null hypothesis because -0.879 lies in the region between +2.576 and -2.576.
C) Reject the null hypothesis because 37% is less than 40%.
D) Do not reject the null hypothesis because 37% lies in the area between 0% and 40%.

Accepted Answer

The sample proportion is 74/200 = 0.37, and the standard error of the proportion is sqrt(0.4 * 0.6 / 200) = 0.044. The test statistic is (0.37 - 0.4) / 0.044 = -0.879. The critical value for a two-tailed test at the 0.02 significance level is 2.326. Since -0.879 lies in the region between +2.326 and -2.326, we do not reject the null hypothesis.

Question 4

A sample of 250 adults tried the new multigrain cereal "Wow!" A total of 187 rated it as excellent. In a sample of 100 children, 66 rated it as excellent. Using the 0.1 significance level, the researcher wishes to show that adults like the cereal better than children. Which of the following is the alternate hypothesis? A) H₁: π_A = π_C B) H₁: π_A < π_C C) H₁: π_A > π_C D) H₁: π_A ≠ π_C

Accepted Answer

The hypothesis should state that adults like the cereal better than children, which means that the proportion of adults who rate it as excellent is greater than the proportion of children who rate it as excellent. Therefore, the alternate hypothesis should be one-tailed and state that πA > πC, where πA is the true proportion of adults who rate the cereal as excellent and πC is the true proportion of children who rate the cereal as excellent. Option C is the only one-tailed hypothesis that fits this requirement.

Question 5

In a market test of a new chocolate raspberry coffee, a poll of 400 people from Dobbs Ferry showed 250 preferred the new coffee. In Irvington, 170 out of 350 people preferred the new coffee. To test the hypothesis that there is no difference in preferences between the two villages, what is the null hypothesis? A) H₀: π₁ < π₂ B) H₀: π₁ > π₂ C) H₀: π₁ = π₂ D) H₀: π₁ ≠ π₂

Accepted Answer

The null hypothesis always assumes that there is no difference between two groups or populations being compared. In this case, the null hypothesis is that there is no difference in preferences between the two villages, which can be represented as the statement π₁ = π₂, where π represents the proportion of people who prefer the new coffee. Therefore, the correct choice is C.

Question 6

How is a pooled estimate of the population proportion represented? A) p_c B) z C) π D) nπ

Accepted Answer

The pooled estimate of the population proportion is represented by $ \hat{p} $, which is often used in hypothesis testing or constructing confidence intervals when comparing two proportions from independent samples. This symbol is not explicitly listed in the options, but option A seems to attempt to represent a special character or formula, which could be interpreted as an attempt to denote $ \hat{p} $ or a related statistical symbol.

Question 7

To test the hypothesis that 55% of those families who plan to purchase a vacation residence in Florida want a condominium, the null hypothesis is π = 0.55 and the alternate is π ≠ .55. A random sample of 400 families who planned to buy a vacation residence revealed that 228 families want a condominium. What decision should be made regarding the null hypothesis using the 0.01 level of significance?
A) Do not reject it.
B) Reject it.
C) Cannot accept it or reject it based on the information given.
D) Purchase a condominium.

Accepted Answer

The sample proportion of families who want a condominium is 228/400 = 0.57. The standard error of the proportion is sqrt(0.55*0.45/400) = 0.024. The z-score is (0.57 - 0.55) / 0.024 = 0.83. The p-value is P(|Z| > 0.83) = 2*P(Z > 0.83) = 2*0.2033 = 0.4066. Since the p-value is greater than the significance level of 0.01, we fail to reject the null hypothesis.

Question 8

An electronics retailer believes that, at most, 40% of their cell phone inventory was sold during November. The retailer surveyed 80 dealers and found that 38% of the inventory was sold. Since 38% is less than 40%, is this difference of 2 percentage points sampling error or a significant difference? Test at the 0.05 level. The computed z = -0.37.
A) The 2% is a significant difference.
B) We cannot determine if the 2% is a significant difference.
C) There is not enough information to reach a conclusion.
D) None Apply.

Accepted Answer

Since the computed z-value is less than the critical value at the 0.05 level, we fail to reject the null hypothesis that the proportion of inventory sold is equal to or greater than 40%. Therefore, the difference of 2 percentage points is likely due to sampling error and not a significant difference.

Question 9

A sample of 250 adults tried the new multigrain cereal "Wow!" A total of 187 rated it as excellent. In a sample of 100 children, 66 rated it as excellent. Using the 0.1 significance level, the researcher wishes to show that adults like the cereal better than children. What test statistic should we use to compare the ratings of adults and children?
A) A z-statistic
B) A right one-tailed test statistic
C) A left one-tailed test statistic
D) A t-statistic

Accepted Answer

Since the sample sizes are relatively large (250 for adults and 100 for children), we can use a z-test to compare the two proportions. A t-test is typically used when sample sizes are small or when the population standard deviation is unknown. The researcher wants to show that adults like the cereal better than children, so we would use a right one-tailed test to test this hypothesis.

Question 10

Suppose we test H₀: π₁ = π₂ at the 0.05 level of significance. If the z-test statistic is -1.07, what is our decision? A) Reject the null hypothesis. B) Do not reject the null hypothesis. C) Take a larger sample. D) Reserve judgment.

Accepted Answer

The z-test statistic is negative, which indicates the sample proportion for the two groups being compared is slightly smaller in the first group (H₁) than the second group (H₂). However, the calculated p-value corresponding to this test statistic is greater than 0.05, which means that the observed difference between the sample proportions could have occurred by chance under the null hypothesis. Therefore, we do not have sufficient evidence to reject the null hypothesis. The correct answer is B, do not reject the null hypothesis. Taking a larger sample would not change this result since the p-value is greater than 0.05. Therefore, C is not the correct answer. Additionally, since we have a calculated test statistic and corresponding p-value, there is no need to reserve judgment, so D is not the correct answer.

Question 11

A frequency distribution has a mean of 200 and a standard deviation of 20. The class limits for one class are 220 up to 240. Based on the normal distribution, what is the probability that an observation would be in this class?
A) 0.1359
B) 0.3413
C) 0.4772
D) 0.8185

Accepted Answer

The class limits 220 to 240 are 1 and 2 standard deviations above the mean, respectively. Using the empirical rule, approximately 34.13% of data falls between 1 standard deviation above and below the mean, and 13.59% falls between 2 and 3 standard deviations above the mean. Since the class is between 1 and 2 standard deviations above the mean, the probability is 34.13% - 13.59% = 13.59% for the upper half.

Question 12

A sample of 250 adults tried the new multigrain cereal "Wow!" A total of 187 rated it as excellent. In a sample of 100 children, 66 rated it as excellent. Using the 0.1 significance level, the researcher wishes to show that adults like the cereal better than children. What is the pooled proportion?
A) 0.723
B) 1.408
C) 0.494
D) 0.807

Accepted Answer

The pooled proportion is calculated by taking the total number of successes (excellent ratings) divided by the total number of trials (both adults and children). Here, (187 adults + 66 children) / (250 adults + 100 children) = 253 / 350 = 0.723.

Question 13

The sample proportion is defined as _________.
A) nπ
B) x/n
C) n!
D) π

Accepted Answer

The sample proportion is defined as the number of individuals in a sample with a certain characteristic (x) divided by the total sample size (n), which is represented as x/n.

Question 14

In a market test of a new chocolate raspberry coffee, a poll of 400 people (sample 1) from Dobbs Ferry showed 250 preferred the new coffee. In Irvington, 170 out of 350 people (sample 2) preferred the new coffee. To test the hypothesis that a higher proportion of people in Dobbs Ferry prefer the new coffee, what is the alternate hypothesis? A) H₁: π₁ < π₂ B) H₁: π₁ > π₂ C) H₁: π₁ = π₂ D) H₁: π₁ ≠ π₂

Accepted Answer

The alternate hypothesis should state that the proportion of people who prefer the new coffee in Dobbs Ferry is higher than the proportion in Irvington. Option B states "π₁ > π₂", which means the proportion of people who prefer the new coffee in Dobbs Ferry (sample 1) is greater than the proportion in Irvington (sample 2). This aligns with the hypothesis we want to test.

Question 15

A frequency distribution has a mean of 200 and a standard deviation of 20. The class limits for one class are 220 up to 240. What are the standard normal z-statistics for the class limits?
A) -20 and 20
B) -2.0 and -1.0
C) 200 and 220
D) 1.0 and 2.0

Accepted Answer

The z-statistic is calculated using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For the lower limit (220), $z = \frac{(220 - 200)}{20} = 1.0$. For the upper limit (240), $z = \frac{(240 - 200)}{20} = 2.0$.

Question 16

The number of trials and the population proportion are respectively represented by what symbols?
A) p and n
B) α and β
C) z and t
D) n and π

Accepted Answer

The symbol n represents the number of trials or sample size, while the symbol π represents the population proportion. Therefore, option D is the correct answer. Option A is incorrect because p typically represents the sample proportion, not the number of trials. Option B represents the level of significance and the probability of Type II error in hypothesis testing, while option C represents statistical parameters such as the z-score and t-score.

Question 17

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 null hypothesis? A) H₀: π = 0.36 B) H₀: π = 0.41 C) H₀: π ≠ 0.36 D) H₀: µ = 0.41

Accepted Answer

Null hypothesis always assumes that there is no significant difference between the observed and expected values. Here, the local CBS affiliate claims that their late evening news reaches 41% of the viewing audience, which is the expected value. On the other hand, the survey of 100 viewers shows that 36% of the viewers watch the late evening news on this local CBS station, which is the observed value. Therefore, the null hypothesis should be that the proportion of viewers who watch the late evening news on this local CBS station is equal to the claimed proportion provided by the CBS affiliate. Hence, the correct null hypothesis is H₀: π = 0.41 (Option B).

Question 18

A frequency distribution has a mean of 100 and a standard deviation of 20. The class limits for one class are 50 up to 60. What are the standard normal z-statistics for the class limits?
A) -20 and 20
B) -2.5 and -2.0
C) 2.0 and 2.5
D) -50 and -40

Accepted Answer

To calculate the z-statistics we use the formula:

z = (x - μ) / σ

For the lower class limit x = 50, μ = 100, and σ = 20, we get:

z = (50 - 100) / 20 = -2.5

For the upper class limit x = 60, μ = 100, and σ = 20, we get:

z = (60 - 100) / 20 = -2.0

Therefore, the standard normal z-statistics for the class limits are -2.5 and -2.0.

Question 19

A frequency distribution has a mean of 100 and a standard deviation of 20. The class limits for one class are 50 up to 60. Based on the normal distribution, what is the probability that an observation would be in this class?
A) 0.4938
B) 0.4772
C) 0.0166
D) -0.0166

Accepted Answer

To find the probability of an observation being in this class, we need to standardize the values using the formula:
z = (x - mean) / standard deviation
For the lower limit of the class, z = (50 - 100) / 20 = -2.5
For the upper limit of the class, z = (60 - 100) / 20 = -2
Using a standard normal distribution table, the probability of z being between -2.5 and -2 is 0.0166. Therefore, the probability of an observation being in this class is 0.0166.

Question 20

To test if an observed frequency distribution with five classes is normally distributed, we compute probabilities for each class based on a(n) _________________.
A) Standard normal distribution
B) Chi-square distribution
C) Student's t distribution
D) F distribution

Accepted Answer

The standard normal distribution is used to measure probabilities of normally distributed data, which includes testing for normality of a frequency distribution with five classes.

Quiz 15: Nonparametric Methods: Nominal Level Hypothesis Tests

If 20 out of 50 students sampled live in a college dormitory, what is the estimated proportion of students at the University living in a dormitory?

If the decision is to reject the null hypothesis of no difference between two population proportions at the 5% level of significance, what are the alternative hypothesis and rejection region?

The claim that "40% of those persons who retired from an industrial job before the age of 60 would return to work if a suitable job was available" is to be investigated at the 0.02 significance level. If 74 out of the 200 workers sampled said they would return to work, what is our decision?

How is a pooled estimate of the population proportion represented?

Suppose we test H0: π1 = π2 at the 0.05 level of significance. If the z-test statistic is -1.07, what is our decision?

A frequency distribution has a mean of 200 and a standard deviation of 20. The class limits for one class are 220 up to 240. Based on the normal distribution, what is the probability that an observation would be in this class?

A sample of 250 adults tried the new multigrain cereal "Wow!" A total of 187 rated it as excellent. In a sample of 100 children, 66 rated it as excellent. Using the 0.1 significance level, the researcher wishes to show that adults like the cereal better than children. What is the pooled proportion?

The sample proportion is defined as _________.

A frequency distribution has a mean of 200 and a standard deviation of 20. The class limits for one class are 220 up to 240. What are the standard normal z-statistics for the class limits?

The number of trials and the population proportion are respectively represented by what symbols?

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 null hypothesis?

A frequency distribution has a mean of 100 and a standard deviation of 20. The class limits for one class are 50 up to 60. What are the standard normal z-statistics for the class limits?

A frequency distribution has a mean of 100 and a standard deviation of 20. The class limits for one class are 50 up to 60. Based on the normal distribution, what is the probability that an observation would be in this class?

To test if an observed frequency distribution with five classes is normally distributed, we compute probabilities for each class based on a(n) _________________.

Suppose we test H₀: π₁ = π₂ at the 0.05 level of significance. If the z-test statistic is -1.07, what is our decision?