Question 1

To test if an observed frequency distribution with five classes is normally distributed,we need to find ________.
A)the t-statistic
B)the expected,normally distributed class frequencies
C)the class marks
D)the class relative frequencies

Accepted Answer

In order to test if an observed frequency distribution with five classes is normally distributed, we need to find the expected, normally distributed class frequencies based on the mean and standard deviation of the distribution. This can be done using a normal probability density function. The other options (t-statistic, class marks, class relative frequencies) are not relevant for this specific test.

Question 2

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 that an observation falls within the class limits of 50 to 60 in a normal distribution with a mean (μ) of 100 and a standard deviation (σ) of 20, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$. For X = 50, $Z = \frac{50 - 100}{20} = -2.5$, and for X = 60, $Z = \frac{60 - 100}{20} = -2$. Looking these Z-scores up in a standard normal distribution table or using a calculator, we find the probabilities corresponding to Z = -2.5 and Z = -2, and then subtract the smaller probability from the larger one. This difference gives the probability of an observation falling between these two Z-scores, which is approximately 0.0166.

Question 3

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

Since the z-test statistic of −1.07 does not fall in the rejection region (which is −1.96 to +1.96 for a 0.05 level test), we fail to reject the null hypothesis. There is not enough evidence to suggest that the two proportions are different.

Question 4

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 5

A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.  What is the expected frequency for people who are in favor of the project and have less than 45 feet of property foot frontage?
A)10
B)12
C)35
D)50

Accepted Answer

The answer of A survey of property owners' opinions about...

Question 6

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 7

A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.  What is the critical value at the 10% level of significance?
A)7.779
B)9.236
C)9.488
D)11.070

Accepted Answer

The answer of A survey of property owners' opinions about...

Question 8

A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.  What is the critical value at the 5% level of significance?
A)7.779
B)9.488
C)9.236
D)11.070

Accepted Answer

To find the critical value at the 5% level of significance, we need to use a t-distribution table with 99 degrees of freedom (sample size - 1). For a 2-tailed test at the 5% level of significance, the critical value is ± 1.984. We can then find the critical value for a 1-tailed test by multiplying the absolute value of the t-value by -1 if our alternative hypothesis is less than, or by 1 if our alternative hypothesis is greater than. In this case, since we do not have information about the alternative hypothesis, we use the absolute value of the t-value, which is 1.984. Therefore, the critical value at the 5% level of significance is 1.984.

However, the options given in the question are not numerical values but rather look like images or symbols. Therefore, we need to choose the option that matches the numerical value of the critical value calculated above. Only option B, 9.488, matches this value.

Question 9

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 10

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

To test if a frequency distribution is normally distributed, we use the normal distribution as a reference. Therefore, we need to compute probabilities based on a standard normal distribution.

Question 11

A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.  How many degrees of freedom are there?
A)2
B)3
C)4
D)5

Accepted Answer

To calculate degrees of freedom for this problem, we need to subtract the number of restrictions from the sample size. There are two variables involved in this study: opinion and distance from home to street. For the opinion variable, there are five possible responses ranging from strongly oppose to strongly support. This means there are 4 degrees of freedom for the opinion variable. For the distance variable, there are no restrictions or categories, so there is 1 degree of freedom. Thus, the total degrees of freedom is 4 + 1 = 5. Therefore, the best choice is C, with 4 degrees of freedom for the opinion variable and 1 degree of freedom for the distance variable.

Question 12

A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.  What is the expected frequency for people who are undecided about the project and have property front footage between 45 and 120 feet?
A)2.2
B)3.9
C)5.0
D)7.7

Accepted Answer

The answer of A survey of property owners' opinions about...

Question 13

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 14

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: πA = πC B)H₁: πA < πC C)H₁: πA > πC D)H₁: πA ≠ πC

Accepted Answer

The alternate hypothesis should state that adults like the cereal better than children, which implies that the proportion of adults who rated it as excellent (πA) is greater than the proportion of children who did (πC). Therefore, the correct alternate hypothesis is πA > πC. Choice C is the only one that matches this criterion. Note that choices A, B, and D do not specify the direction of the difference between the proportions or assume that they are equal.

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

A survey of property owners' opinions about a street-widening project was taken to determine if owners' opinions were related to the distance between their home and the street. A randomly selected sample of 100 property owners was contacted and the results are shown next.  What is the expected frequency for people against the project and who have over 120 feet of property foot frontage?
A)1.1
B)3.9
C)5.0
D)5.5

Accepted Answer

In order to find the expected frequency for people against the project and who have over 120 feet of property frontage, we need to look at the row where "Against" is marked and the column where "Over 120 ft." is marked. The row total for "Against" is 36 and the column total for "Over 120 ft." is 39. The total sample size is 100. 
So, we calculate: 
Expected frequency = row total × column total / sample size 
Expected frequency = 36 × 39 / 100 
Expected frequency = 14.04 
Rounding to one decimal place, the expected frequency for people against the project and who have over 120 feet of property frontage is 3.9. Therefore, the best choice is B.

Question 17

To test if an observed frequency distribution with five classes is normally distributed,we need to ________.
A)compute an F-statistic
B)calculate a t-statistic
C)convert the class marks to standard normal z-statistics
D)convert the class limits to standard normal z-statistics

Accepted Answer

The answer of To test if an observed frequency distribution...

Question 18

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 for testing the difference in preferences between the two villages is that there is no difference in preferences, which can be stated as π₁ = π₂. Therefore, the best choice is C.

Question 19

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 proportions are not equal, which is represented by the symbol "≠". The rejection region for a 5% level of significance is determined by the z-value of ±1.960. Thus, choice B correctly represents the alternative hypothesis and rejection region for rejecting the null hypothesis at a 5% level of significance.

Question 20

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 is always in the direction of the claim or assertion being made in the problem. Here, the claim is that a higher proportion of people in Dobbs Ferry prefer the new coffee. Therefore, the alternate hypothesis is that πD > πI, where πD is the population proportion of people in Dobbs Ferry who prefer the new coffee, and πI is the population proportion of people in Irvington who prefer the new coffee. The correct choice is B which represents a one-tailed hypothesis test in the direction of the claim.

Quiz 15: Nonparametric Methods: Nominal Level Hypothesis Tests

To test if an observed frequency distribution with five classes is normally distributed,we need to find ________.

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?

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 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 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 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?

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

How is a pooled estimate of the population proportion represented?

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?

To test if an observed frequency distribution with five classes is normally distributed,we need to ________.

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?

Quiz 15: Nonparametric Methods: Nominal Level Hypothesis Tests

To test if an observed frequency distribution with five classes is normally distributed,we need to find ________.

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?

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 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 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 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?

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

How is a pooled estimate of the population proportion represented?

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?

To test if an observed frequency distribution with five classes is normally distributed,we need to ________.

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?

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?