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The proportion of voters each of the five parties:
|
Party |
Bloc Quebecois |
Conservative |
Green Party |
Liberal Party |
New Democratic |
||||||
|
Proportion |
0.06 |
0.40 |
0.04 |
0.19 |
0.31 |
A sample of 500 voters was interviewed one year after the election and asked which party they support.
The results are shown in the table below:
|
Count |
37 |
171 |
25 |
119 |
148 |
||||
|
Expected |
|
||||||||
|
Cell Chi-Square |
We would like to conduct a chi-square goodness of fit test at the 10% level of significance to determine whether party support has changed in Canada since the last election.
(a) Calculate the expected cell counts and the cell chi-square values and enter them in the table above. (Round all values to
two
decimal places.)
(b) There are no expected cell counts less than 5, so we can safely use the chi-square distribution.
The value of the test statistic for the appropriate test of significance (rounded to two decimal places) is
.
(c) The P-value for the appropriate test of significance is between and .
(d) What is the correct conclusion of the test?
Reject H0. There is insufficient evidence that party support has changed since the election.
Reject H0. There is insufficient evidence that party support has remained the same since the election.
Fail to reject H0. There is sufficient evidence that party support has remained the same since the election.
Fail to reject H0. There is sufficient evidence that party support has changed since the election.
Fail to reject H0. There is insufficient evidence that party support has changed since the election.
Reject H0. There is sufficient evidence that party support has remained the same since the election.
Fail to reject H0. There is insufficient evidence that party support has remained the same since the election.
Reject H0. There is sufficient evidence that party support has changed since the election.
(e) If we used the critical value method to conduct the test, the decision rule would be to reject H0 if
Each player is dealt a hand of five cards from a standard deck of
5
2
cards. One player records the number of Hearts in each of the
1
7
0
hands he plays one night. The data are shown in the table below:
|
# of Hearts |
0 | 1 | 2 |
3 |
4 |
5 | ||||||
|
# of Hands |
33 |
72 |
49 |
13 |
||||||||
|
Expected |
|
We would like to conduct a chi-square goodness of fit test at the 10% level of significance to determine whether the number of Hearts per hand follows a binomial distribution with parameter p = 0.25.
(a) Calculate all expected cell counts and enter them in the table above. (Round all values to
two
decimal places.)
(b) The expected cell counts for 4 and 5 Hearts are both less than 5, so we must merge them with the cell for 3 Hearts. Enter all appropriate expected counts in the table below. Then calculate the cell chi-square values and enter them in the table. (Round all values to
two
decimal places.)
|
16 |
|||
|
Cell Chi-Square |
(c) The value of the test statistic for the appropriate test of significance (rounded to
two
decimal places) is
(d) The P-value for the appropriate test of significance is:
between 0.01 and 0.02.
between 0.02 and 0.025.
between 0.025 and 0.05.
between 0.05 and 0.10.
between 0.10 and 0.15.
between 0.15 and 0.20.
between 0.20 and 0.25.
greater than 0.25.
(e) Suppose we had instead used the critical value method to conduct the test. The decision rule would be to reject H0 if .
(f) What is the correct conclusion for this test?
Fail to reject H0. There is insufficient evidence that the number of Hearts per hand does not follow a binomial distribution with parameter p = 0.25.
Fail to reject H0. There is sufficient evidence that the number of Hearts per hand follows a binomial distribution with parameter p = 0.25.
Reject H0. There is insufficient evidence that the number of Hearts per hand does not follow a binomial distribution with parameter p = 0.25.
Fail to reject H0. There is sufficient evidence that the number of Hearts per hand does not follow a binomial distribution with parameter p = 0.25.
Fail to reject H0. There is insufficient evidence that the number of Hearts per hand follows a binomial distribution with parameter p = 0.25.
Reject H0. There is sufficient evidence that the number of Hearts per hand does not follow a binomial distribution with parameter p = 0.25.
Reject H0. There is sufficient evidence that the number of Hearts per hand follows a binomial distribution with parameter p = 0.25.
Reject H0. There is insufficient evidence that the number of Hearts per hand follows a binomial distribution with parameter p = 0.25.
(g) Which of the following statements must be true?
We have made a Type I Error in our conclusion.
We have made a Type II Error in our conclusion.
We have not made a Type I or a Type II Error. Our conclusion is correct.
It is impossible to know whether we have made an error in our conclusion.
(h) Suppose we only wanted to test whether the number of Hearts per hand follows a binomial distribution (without specifying a particular value of the parameter p). The estimated value of the parameter p (rounded to
four
decimal places) would be .
(i) If the number of Hearts had to be estimated, then, using the critical value method, the decision rule would be to reject H0 if
The number of goals per game scored by teams in a professional soccer league are shown below for a sample of
1
8
0
games:
|
# of Goals |
0 | 1 |
2 |
3 |
4 |
5 |
|||||||
|
# of Games |
57 |
53 |
41 |
19 |
8 |
We would like to conduct a chi-square goodness of fit test at the 5% level of significance to determine whether the number of goals scored per game by teams in the league has a Poisson distribution. Since a Poisson variable can take any non-negative integer value, it is possible that the number of goals is even greater than 5 (even though there are no games in this sample for which X > 5). As such, we must label the last column “”.
|
57 |
53 |
41 |
19 |
8 |
2 |
|
|
Expected Count |
|
(a) Since we don’t know the value of the parameter , we must estimate it. The estimated value of is
.
(b) Enter the expected cell counts in the table above. (When you are calculating probabilities for the Poisson distribution, round your probabilities to four decimal places. Then round your expected cell counts to two decimal places.)
Since the last cell has an expected count less than 5, we must merge it with the cell to its left. We will label this new cell “”. Our final merged table is shown below:
# of Goals
0
1
2
3
# of Games
57
53
41
19
10
Expected
Cell Chi-Square
(c) Enter all expected counts and cell chi-square values (rounded to two decimal places).
(d) The value of the test statistic for the appropriate test of significance (rounded to two decimal places) is .
(e) The P-value for the appropriate test of significance is between and .
(f) If we used the critical value method to conduct the test, the decision rule would be to reject H0 if
.
(g) What is the correct conclusion for this test?
Fail to reject H0. There is insufficient evidence that the number of goals per game by a team does not have a Poisson distribution.
Reject H0. There is insufficient evidence that the number of goals per game by a team does not have a Poisson distribution.
Fail to reject H0. There is sufficient evidence that the number of goals per game by a team does not have a Poisson distribution.
Reject H0. There is insufficient evidence that the number of goals per game by a team has a Poisson distribution.
Fail to reject H0. There is insufficient evidence that the number of goals per game by a team has a Poisson distribution.
Reject H0. There is sufficient evidence that the number of goals per game by a team has a Poisson distribution.
Reject H0. There is sufficient evidence that the number of goals per game by a team does not have a Poisson distribution.
Fail to reject H0. There is sufficient evidence that the number of goals per game by a team has a Poisson distribution.
*Round all answers to
two decimal places, unless otherwise noted.
We would like to determine how the time a student spends studying for an exam affects his or her score. Study times (in hours) and exam scores (in %) are shown in the table below for a sample of eight students:
Study Time
11
2
3
5
14
17
2
8
Score
50
94
41
82
62
73
85
64
From these data, we calculate , , sx = 8.72, sy = 18.07, r = 0.795, and .
(a) The equation of the least squares regression line is + x.
(b) What percentage of the variation in Exam Score is accounted for by its regression on Study Time? %
(c) What is the predicted Exam Score for a student who studies 10 hours?
(d) What is the value of the residual for the student in the sample who studied for 17 hours?
(e) The estimate of the parameter in the regression model is = .
(f) A 95% confidence interval for the mean exam score of all students who study 12 hours is
( , ).
(g) A 95% prediction interval for the final exam score for a student who studies 12 hours is
( , ).
(h) A 95% confidence interval for the slope of the true regression line is ( , ).
(i) We would like to conduct a hypothesis test of H0: = 0 vs. Ha: 0 at the 5% level of significance. The value of the test statistic for the appropriate test of significance is t = .
(j) The P-value of the appropriate test of significance is between and .
(k) Use JMP to find the exact P-value of the test. The exact P-value (rounded to four decimal places) is .
(l) Using only the result in (i), what would be the value of the test statistic (rounded to two decimal places) if we had conducted this test using an analysis of variance? F =
(m) Using only the result in (k), what would be the P-value of the test (rounded to four decimal places) if we had conducted this test using an analysis of variance?
(n) What is the correct conclusion of the test?
Reject H0. There is sufficient evidence that Study Time and Exam Score do not a linear relationship.
Fail to reject H0. There is insufficient evidence that Study Time and Exam Score do not have a linear relationship.
Reject H0. There is sufficient evidence that Study Time and Exam Score have a linear relationship.
Fail to reject H0. There is sufficient evidence that Study Time and Exam Score have a linear relationship.
A sample of 17 breakfast cereals is selected and the amount of sugar and the amount of sodium per serving (both measured in grams) are recorded. The correlation between sugar and sodium content for these cereals is calculated to be -0.434.
We would like to conduct a hypothesis test at the 1% level of significance to determine whether there exists a negative correlation between the amount of sugar and the amount of sodium in the population of all breakfast cereals.
(a) The hypotheses for the appropriate test of significance are:
|
H0: vs. Ha: |
|
H0: r = 0 vs. Ha: r > 0 |
|
H0: r = 0 vs. Ha: r < 0 |
H0: vs. Ha:
H0: vs. Ha:
H0: vs. Ha:
H0: vs. Ha:
H0: vs. Ha:
(b) The value of the test statistic for the appropriate test of significance (rounded to
two
decimal places) is
t = .
(c) The P-value for the appropriate test of significance is between and .
(d) If we conducted the test using the critical value method, the decision rule would be to reject H0 if
.
(e) What is the correct conclusion for this test?
Reject H0. There is insufficient evidence that there exists a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Fail to reject H0. There is insufficient evidence that there does not exist a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Fail to reject H0. There is sufficient evidence that there exists a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Reject H0. There is sufficient evidence that there does not exist a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Reject H0. There is insufficient evidence that there does not exist a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Fail to reject H0. There is insufficient evidence that there exists a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Reject H0. There is sufficient evidence that there exists a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Fail to reject H0. There is sufficient evidence that there does not exist a negative correlation between the amount of sugar and the amount of sodium in breakfast cereals.
Each year, the Federal Trade Commission tests the tar and
nicotine content of various brands of cigarettes in the United States. Data for a sample of eight brands is collected, and we would like to conduct a hypothesis test at the 5% level of significance to determine whether the tar content of cigarettes can be predicted by their nicotine content. All measurements are in milligrams. The data are analyzed using an analysis of variance, and the ANOVA table (with some values missing) is shown below:
|
Source of Variation |
df |
Sum of Squares |
Mean Squares |
F |
||||
|
Regression |
|
59.04 |
||||||
|
Error |
||||||||
|
Total |
69.90 |
(a) Fill in all of the missing values in the table.
(b) What is the estimate of the parameter in the simple linear regression model? = (Round your answer to
two
decimal places.)
(c) What proportion of the variation in Tar Content can be accounted for by its regression on Nicotine Content? (Round your answer to
two
decimal places.)
(d) We conduct a hypothesis test of H0: vs. Ha: to determine whether there exists a linear relationship between Nicotine Content and Tar Content. The P-value of the appropriate test of significance is between and .
(e) If we used the critical value method to conduct the test, the decision rule would be to reject H0 if
