Question 1

Suppose that a random sample of 150 observations was drawn from a population. After calculating the mean and standard deviation, each observation was standardized and the number of observations in each of the intervals below was counted. Can we infer at the 5% significance level that the data were drawn from a normal population?
 $$\begin{array} { | l | c| } 
\hline \text { Intervals } & \text { Frequency } \
\hline Z \leq - 1.5 & 15 \
- 1.5 < Z \leq - .5 & 32 \
- .5 \leq Z \leq .5 & 65 \
.5 < Z \leq 1.5 & 25 \
Z > 1.5 & 13 \
\hline
\end{array}$$

Accepted Answer

H₀: p₁ = 0.0668, p₂ = 0.2417, p₃ = 0.3830, p₄ = 0.2417, p₅ = 0.0668 (The population is normal) H₁: At least two proportions differ from their specified values (The population is not normal) Rejection region: $\chi$² > $\chi$²_0.05,2 = 5.991 Test statistic: $\chi$² = 8.347 Conclusion: Reject the null hypothesis. We can't infer at the 5% significance level that the data were drawn from a normal population.

Question 2

We can use the goodness-of-fit test to determine whether data were drawn from any distribution of interest. The most common application of this procedure is a test of ____________________.

Accepted Answer

normality

Question 3

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

Accepted Answer

null
H<sub>0</sub>

Question 4

To test for normality, the ____________________ hypothesis is that at least two proportions differ from their specified values.

Accepted Answer

The answer of To test for normality, the ____________________ hypothesis...

Question 5

A large value of the chi-squared test statistic in a test of normality means you reject H₀ and conclude that the data ____________________ (do/do not) come from a normal distribution.

Accepted Answer

The answer of A large value of the chi-squared test...

Question 6

The number of degrees of freedom associated with the chi-squared test statistic for normality is the number of ____________________ minus 1 minus the number of ____________________ estimated.

Accepted Answer

The answer of The number of degrees of freedom associated...

Question 7

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 200 observations. In addition, the researcher used 5 standardized intervals to test for normality. Using a 10% level of significance, the critical value for this test is 4.60517.

Accepted Answer

This statement is true. The critical value for a goodness-of-fit test with 5 intervals and a significance level of 10% is indeed 4.60517, assuming the null hypothesis is true (i.e., the data comes from a normally distributed population).

Question 8

The chi-squared test for normality must follow the rule of ____________________ regarding expected values.

Accepted Answer

The answer of The chi-squared test for normality must follow...

Question 9

The number of degrees of freedom in a chi-squared test for normality, where the number of standardized intervals is 5 and there are 2 population parameters to be estimated from the data, is equal to:
A)5
B)4
C)3
D)2

Accepted Answer

The number of degrees of freedom is calculated by subtracting the number of estimated parameters from the number of standardized intervals. In this case, 5-2=3, so the answer is choice D.

Question 10

The null hypothesis states that the sample data came from a normally distributed population. The researcher calculates the sample mean and the sample standard deviation from the data. The data arrangement consisted of five categories. Using $\alpha$ = 0.05, the appropriate critical value for this chi-squared test for normality is 5.99147.

Accepted Answer

The critical value for a chi-squared test for normality with 5 categories and $\alpha$ = 0.05 is 5.99147.

Question 11

If we want to perform a one-tail test of a population proportion p, we can employ either the z-test of p, or the chi-squared goodness-of-fit test.

Accepted Answer

The answer of If we want to perform a one-tail...

Question 12

The number of degrees of freedom in testing for normality is the:
A)number of intervals used to test the hypothesis minus one.
B)number of parameters estimated minus one.
C)number of intervals used to test the hypothesis minus one minus the number of parameters estimated.
D)None of these choices.

Accepted Answer

The number of degrees of freedom in testing for normality is calculated by taking the number of intervals used to test the hypothesis, subtracting one, and then subtracting the number of parameters estimated. This accounts for the constraints imposed by the estimation of parameters and the partitioning of data into intervals.

Question 13

Suppose that a random sample of 60 observations was drawn from a population. After calculating the mean and standard deviation, each observation was standardized and the number of observations in each of the intervals below was counted. Can we infer at the 10% significance level that the data were drawn from a normal population?
 $$\begin{array} { | l | c | } 
\hline \text { Intervals } & \text { Frequency } \
\hline Z \leq - 1 & \mathbf { 8 } \
- 1 < Z \leq 0 & 30 \
0 < Z \leq 1 & 17 \
Z > 1 & 5 \
\hline
\end{array}$$

Accepted Answer

The answer of Suppose that a random sample of 60...

Question 14

If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of p₁-p₂, or the chi-squared test of a contingency table.

Accepted Answer

Both z-test and chi-squared test can be used to test for differences between two populations of nominal data with exactly two categories.

Question 15

If we want to perform a two-tail test of a population proportion p, we can only use the z-test of p.

Accepted Answer

We can also use the chi-square test or the t-test for the two-tail test of a population proportion. The choice of test depends on the sample size and other factors.

Question 16

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 500 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a 5% level of significance, the critical value for this test is:
A)11.1433
B)9.3484
C)7.8147
D)9.4877

Accepted Answer

The critical value for a goodness-of-fit test with 6 intervals and a significance level of 5% is 7.8147. This can be found using a chi-squared distribution table with 6-1=5 degrees of freedom and a significance level of 0.05. The researcher's estimate of the population mean and standard deviation are not relevant to determining the critical value in this scenario.

Question 17

In a goodness-of-fit test, the null hypothesis states that the data came from a normally distributed population. The researcher estimated the population mean and population standard deviation from a sample of 300 observations. In addition, the researcher used 6 standardized intervals to test for normality. Using a 2.5% level of significance, the critical value for this test is 14.4494.

Accepted Answer

The critical value depends on the degrees of freedom, which in this case is 6 - 2 = 4. Using a significance level of 2.5%, the correct critical value is actually 9.4877.

Question 18

When the problem objective is to describe a population of nominal data with exactly two categories, we can employ either the z-test of a population proportion p, or the chi-squared goodness-of-fit test.

Accepted Answer

Both the z-test of a population proportion p and chi-squared goodness-of-fit test can be used to describe a population of nominal data with exactly two categories. The z-test compares a sample proportion to a hypothesized population proportion, while the chi-squared goodness-of-fit test compares observed frequencies to expected frequencies assuming a certain proportion.

Question 19

The following data are believed to have come from a normal probability distribution. $$\begin{array} { l l l l l l l l l } 
26 & 21 & 25 & 20 & 21 & 29 & 26 & 23 & 22 \
24 & 30 & 23 & 32 & 26 & 24 & 32 & 16 & 36 \
21 & 31 & 26 & 23 & 32 & 35 & 40 & 30 & 14 \
46 & 27 & 33 & 25 & 27 & 21 & 26 & 18 & 29
\end{array}$$ The mean of this sample equals 26.80, and the standard deviation equals 6.378. Use the goodness-of-fit test at the 5% significance level to test whether the data indeed come from a normal distribution.

Accepted Answer

The answer of The following data are believed to have...

Question 20

The number of degrees of freedom associated with the chi-squared test for normality is the number of intervals used minus the number of parameters estimated from the data.

Accepted Answer

The answer of The number of degrees of freedom associated...

Quiz 15: Chi-Squared Tests Optional

We can use the goodness-of-fit test to determine whether data were drawn from any distribution of interest. The most common application of this procedure is a test of ____________________.

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

To test for normality, the ____________________ hypothesis is that at least two proportions differ from their specified values.

A large value of the chi-squared test statistic in a test of normality means you reject H₀ and conclude that the data ____________________ (do/do not) come from a normal distribution.

The number of degrees of freedom associated with the chi-squared test statistic for normality is the number of minus 1 minus the number of estimated.

The chi-squared test for normality must follow the rule of ____________________ regarding expected values.

The number of degrees of freedom in a chi-squared test for normality, where the number of standardized intervals is 5 and there are 2 population parameters to be estimated from the data, is equal to:

If we want to perform a one-tail test of a population proportion p, we can employ either the z-test of p, or the chi-squared goodness-of-fit test.

The number of degrees of freedom in testing for normality is the:

If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of p₁-p₂, or the chi-squared test of a contingency table.

If we want to perform a two-tail test of a population proportion p, we can only use the z-test of p.

When the problem objective is to describe a population of nominal data with exactly two categories, we can employ either the z-test of a population proportion p, or the chi-squared goodness-of-fit test.

The number of degrees of freedom associated with the chi-squared test for normality is the number of intervals used minus the number of parameters estimated from the data.

Quiz 15: Chi-Squared Tests Optional

We can use the goodness-of-fit test to determine whether data were drawn from any distribution of interest. The most common application of this procedure is a test of ____________________.

To test for normality, the ____________________ hypothesis specifies probabilities of certain intervals within the normal distribution.

To test for normality, the ____________________ hypothesis is that at least two proportions differ from their specified values.

A large value of the chi-squared test statistic in a test of normality means you reject H0 and conclude that the data ____________________ (do/do not) come from a normal distribution.

The number of degrees of freedom associated with the chi-squared test statistic for normality is the number of ____________________ minus 1 minus the number of ____________________ estimated.

The chi-squared test for normality must follow the rule of ____________________ regarding expected values.

The number of degrees of freedom in a chi-squared test for normality, where the number of standardized intervals is 5 and there are 2 population parameters to be estimated from the data, is equal to:

If we want to perform a one-tail test of a population proportion p, we can employ either the z-test of p, or the chi-squared goodness-of-fit test.

The number of degrees of freedom in testing for normality is the:

If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of p1-p2, or the chi-squared test of a contingency table.

If we want to perform a two-tail test of a population proportion p, we can only use the z-test of p.

When the problem objective is to describe a population of nominal data with exactly two categories, we can employ either the z-test of a population proportion p, or the chi-squared goodness-of-fit test.

The number of degrees of freedom associated with the chi-squared test for normality is the number of intervals used minus the number of parameters estimated from the data.

A large value of the chi-squared test statistic in a test of normality means you reject H₀ and conclude that the data ____________________ (do/do not) come from a normal distribution.

The number of degrees of freedom associated with the chi-squared test statistic for normality is the number of minus 1 minus the number of estimated.

If we want to test for differences between two populations of nominal data with exactly two categories, we can employ either the z-test of p₁-p₂, or the chi-squared test of a contingency table.