Question 7

A recent newspaper article claims that 56% of car accidents involve some evidence of distracted driving. A police officer questions this claim, and pulls a random sample of 50 accident reports. If 30 of them involve distracted driving, is this enough evidence to show that the proportion of car accidents involve some evidence of distracted driving issignificantlygreater than than reported by the newspaper? (Use=0.10)

1. For this study, we should use
2. Select an answer
3. 1-PropZInt
4. 2-SampTInt
5. 2-SampTTest
6. -Test
7. T-Test
8. TInterval
9. GOF-Test
10. 1-PropZTest
11. ANOVA
12. 2-PropZInt
13. 2-PropZTest

1. The null and alternative hypotheses would be:
2. H0H0:
3. ?
5. p
7. Select an answer
8. >
10. =
11. <
12. (please enter a decimal)
13. H1H1:
14. ?
15. p
18. Select an answer
19. <
20. >
22. =
23. (Please enter a decimal)

1. The test statistic =(please show your answer to 3 decimal places.)

1. The p-value =(Please show your answer to 4 decimal places.)

1. The p-value is
2. ?
4. >

1. Based on this, we should
2. Select an answer
3. accept
4. fail to reject
5. reject
6. the null hypothesis.

1. As such, the final conclusion is that ...

• The sample data suggest that the populaton proportion issignificantlygreater than 56% at= 0.10, so there is sufficient evidence to conclude that the proportion of car accidents involve some evidence of distracted driving is greater than 56%
• The sample data suggest that the population proportion is notsignificantlygreater than 56% at= 0.10, so there is not sufficient evidence to conclude that the proportion of car accidents involve some evidence of distracted driving is greater than 56%.

Question 8

In a certain school district, it was observed that 33% of the students in the element schools were classified as only children (no siblings).However, in the special program for talented and gifted children, 93 out of 225 students are only children.The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the=0.02=0.02level of significance.

What is the hypothesized population proportion for this test?

p=p=

(Report answer as a decimal accurate to 2 decimal places.Do not report using the percent symbol.)

Based on the statement of this problem, how many tails would this hypothesis test have?

• one-tailed test
• two-tailed test

Choose the correct pair of hypotheses for this situation:

(A)

(B)

(C)

H0:p=0.33H0:p=0.33

Ha:p<0.33Ha:p<0.33

H0:p=0.33H0:p=0.33

Ha:p0.33Ha:p0.33

H0:p=0.33H0:p=0.33

Ha:p>0.33Ha:p>0.33

(D)

(E)

(F)

H0:p=0.413H0:p=0.413

Ha:p<0.413Ha:p<0.413

H0:p=0.413H0:p=0.413

Ha:p0.413Ha:p0.413

H0:p=0.413H0:p=0.413

Ha:p>0.413Ha:p>0.413

(A)

(B)

(C)

(D)

(E)

(F)

Using the normal approximation for the binomial distribution (without the continuity correction), was is the test statistic for this sample based on the sample proportion?

z=z=

(Report answer as a decimal accurate to 3 decimal places.)

You are now ready to calculate the P-value for this sample.

P-value =

(Report answer as a decimal accurate to 4 decimal places.)

This P-value (and test statistic) leads to a decision to...

• reject the null
• accept the null
• fail to reject the null
• reject the alternative

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.
• There is not sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.
• The sample data support the assertion that there is a different proportion of only children in the G&T program.
• There is not sufficient sample evidence to support the assertion that there is a different proportion of only children in the G&T program.

Question 9

In a certain school district, it was observed that 34% of the students in the element schools were classified as only children (no siblings).However, in the special program for talented and gifted children, 115 out of 275 students are only children.The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the=0.01=0.01level of significance.

What is the hypothesized population proportion for this test?

p=p=

(Report answer as a decimal accurate to 2 decimal places.Do not report using the percent symbol.)

Based on the statement of this problem, how many tails would this hypothesis test have?

• one-tailed test
• two-tailed test

Choose the correct pair of hypotheses for this situation:

(A)

(B)

(C)

H0:p=0.34H0:p=0.34

Ha:p<0.34Ha:p<0.34

H0:p=0.34H0:p=0.34

Ha:p0.34Ha:p0.34

H0:p=0.34H0:p=0.34

Ha:p>0.34Ha:p>0.34

(D)

(E)

(F)

H0:p=0.418H0:p=0.418

Ha:p<0.418Ha:p<0.418

H0:p=0.418H0:p=0.418

Ha:p0.418Ha:p0.418

H0:p=0.418H0:p=0.418

Ha:p>0.418Ha:p>0.418

(A)

(B)

(C)

(D)

(E)

(F)

Using the normal approximation for the binomial distribution (without the continuity correction), what is the test statistic for this sample based on the sample proportion?

z=z=

(Report answer as a decimal accurate to 3 decimal places.)

You are now ready to calculate the P-value for this sample.

P-value =

(Report answer as a decimal accurate to 4 decimal places.)

This P-value (and test statistic) leads to a decision to...

• reject the null
• accept the null
• fail to reject the null
• reject the alternative

As such, the final conclusion is that...

There is sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.

There is not sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.

The sample data support the assertion that there is a different proportion of only children in the G&T program.

There is not sufficient sample evidence to support the assertion that there is a different proportion of only children in the G&T program.

Question 10

In a certain school district, it was observed that 31% of the students in the element schools were classified as only children (no siblings).However, in the special program for talented and gifted children, 153 out of 421 students are only children.The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the=0.05=0.05level of significance.

What is the hypothesized population proportion for this test?

p=p=

(Report answer as a decimal accurate to 2 decimal places.Do not report using the percent symbol.)

Based on the statement of this problem, how many tails would this hypothesis test have?

• one-tailed test
• two-tailed test

Choose the correct pair of hypotheses for this situation:

(A)

(B)

(C)

H0:p=0.31H0:p=0.31

Ha:p<0.31Ha:p<0.31

H0:p=0.31H0:p=0.31

Ha:p0.31Ha:p0.31

H0:p=0.31H0:p=0.31

Ha:p>0.31Ha:p>0.31

(D)

(E)

(F)

H0:p=0.363H0:p=0.363

Ha:p<0.363Ha:p<0.363

H0:p=0.363H0:p=0.363

Ha:p0.363Ha:p0.363

H0:p=0.363H0:p=0.363

Ha:p>0.363Ha:p>0.363

(A)

(B)

(C)

(D)

(E)

(F)

Using the normal approximation for the binomial distribution (without the continuity correction), was is the test statistic for this sample based on the sample proportion?

z=z=

(Report answer as a decimal accurate to 3 decimal places.)

You are now ready to calculate the P-value for this sample.

P-value =

(Report answer as a decimal accurate to 4 decimal places.)

This P-value (and test statistic) leads to a decision to...

• reject the null
• accept the null
• fail to reject the null
• reject the alternative

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.
• There is not sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.
• The sample data support the assertion that there is a different proportion of only children in the G&T program.
• There is not sufficient sample evidence to support the assertion that there is a different proportion of only children in the G&T program.

Question 11

The US Department of Energy reported that 46% of homes were heated by natural gas. A random sample of 349homes in Oregonfound that 158were heated by natural gas.Test the claim that proportion of homes inOregon that were heated by natural gas is different than what was reported. Use a 5% significance level.Give answer to at least 4 decimal places.

What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)

H₀:

Select an answer

x

s

p

p

s

?

=

<

>

H₁:

Select an answer

p

s

s

x

p

?

<

=

>

Based on the hypotheses, find the following:

Test Statistic =

p-value =

Based on the above we choose to

Select an answer

Accept the null hypothesis

Accept the alternative hypotheis

Reject the null hypothesis

Fail to reject the null hypothesis

The correct summary would be:

Select an answer

There is not enough evidence to reject the claim

There is not enough evidence to support the claim

There is enough evidence to reject the claim

There is enough evidence to support the claim

that the proportion of homes inOregon that were heated by natural gas is different than what the DOE reported value of 46%.

Question 12

The US Department of Energy reported that 50% of homes in the USA were heated by natural gas. A random sample of 323homes in Oregon found that 122were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different from the US reported proportion. Use a 10% significance level. Give answers accurate to at least 3 decimal places.

What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)

H₀:

Select an answer

p

s

p

x

s

?

=

<

>

H₁:

Select an answer

p

s

x

s

p

?

>

<

=

Based on the hypotheses, find the following:

Test Statistic =

p-value =

Based on the above we choose to

Select an answer

Accept the null hypothesis

Reject the null hypothesis

Accept the alternative hypotheis

Fail to reject the null hypothesis

The correct summary would be:

Select an answer

There is enough evidence to support the claim

There is enough evidence to reject the claim

There is not enough evidence to support the claim

There is not enough evidence to reject the claim

that the percent of homes in Oregon heated by natural gas is different than 50%.

Question 13

16% of all Americans suffer from sleep apnea. A researcher suspects that a different percentage of those who live in the inner city have sleep apnea. Of the 351 people from the inner city surveyed, 49 of them suffered from sleep apnea. What can be concluded at the level of significance of= 0.10?

1. For this study, we should use
2. Select an answer
3. z-test for a population proportion
4. t-test for a population mean
5. The null and alternative hypotheses would be:
6. Ho:
7. ?
8. p
11. Select an answer
12. >
14. <
15. =
16. (please enter a decimal)
17. H₁:
18. ?
19. p
22. Select an answer
23. =
24. <
26. >
27. (Please enter a decimal)
28. The test statistic
29. ?
30. z
31. t
32. =(please show your answer to 3 decimal places.)
33. The p-value =(Please show your answer to 4 decimal places.)
34. The p-value is
35. ?
36. >
39. Based on this, we should
40. Select an answer
41. accept
42. fail to reject
43. reject
44. the null hypothesis.
45. Thus, the final conclusion is that ...

• The data suggest the population proportion is notsignificantlydifferent from 16% at= 0.10, so there is not sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is different from 16%.
• The data suggest the population proportion is notsignificantlydifferent from 16% at= 0.10, so there is sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is equal to 16%.
• The data suggest the populaton proportion issignificantlydifferent from 16% at= 0.10, so there is sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is different from 16%

1. Interpret the p-value in the context of the study.

• If the sample proportion of inner city residents who have sleep apnea is 14% and if another 351 inner city residents are surveyed then there would be a 29.72% chance that we would conclude either fewer than 16% of all inner city residents have sleep apnea or more than 16% of all inner city residents have sleep apnea.
• There is a 29.72% chance of a Type I error.
• If the population proportion of inner city residents who have sleep apnea is 16% and if another 351 inner city residents are surveyed then there would be a 29.72% chance that either fewer than 14% of the 351 inner city residents surveyed have sleep apnea or more than 18% of the 351 inner city residents have sleep apnea.
• There is a 29.72% chance that the percent of all inner city residents have sleep apnea differs from 16%.

1. Interpret the level of significance in the context of the study.

• There is a 10% chance that aliens have secretly taken over the earth and have cleverly disguised themselves as the presidents of each of the countries on earth.
• If the population proportion of inner city residents who have sleep apnea is 16% and if another 351 inner city residents are surveyed then there would be a 10% chance that we would end up falsely concluding that the proportion of all inner city residents who have sleep apnea is different from 16%.
• There is a 10% chance that the proportion of all inner city residents who have sleep apnea is different from 16%.
• If the population proportion of inner city residents who have sleep apnea is different from 16% and if another 351 inner city residents are surveyed then there would be a 10% chance that we would end up falsely concluding that the proportion of all inner city residents who have sleep apnea is equal to 16%.

Question 14

57% of students entering four-year colleges receive a degree within six years. Is this percent different from for students who play intramural sports? 110 of the 208 students who played intramural sports received a degree within six years. What can be concluded at the level of significance of= 0.05?

1. For this study, we should use
2. Select an answer
3. z-test for a population proportion
4. t-test for a population mean
5. The null and alternative hypotheses would be:
6. Ho:
7. ?
9. p
11. Select an answer
12. >
14. <
15. =
16. (please enter a decimal)
17. H₁:
18. ?
20. p
22. Select an answer
23. >
24. =
25. <
27. (Please enter a decimal)
28. The test statistic
29. ?
30. z
31. t
32. =(please show your answer to 3 decimal places.)
33. The p-value =(Please show your answer to 4 decimal places.)
34. The p-value is
35. ?
37. >
39. Based on this, we should
40. Select an answer
41. fail to reject
42. accept
43. reject
44. the null hypothesis.
45. Thus, the final conclusion is that ...

• The data suggest the population proportion is notsignificantlydifferent from 57% at= 0.05, so there is sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is equal to 57%.
• The data suggest the population proportion is notsignificantlydifferent from 57% at= 0.05, so there is not sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is different from 57%.
• The data suggest the populaton proportion issignificantlydifferent from 57% at= 0.05, so there is sufficient evidence to conclude that the population proportion of students who played intramural sports who received a degree within six years is different from 57%

1. Interpret the p-value in the context of the study.

• There is a 23.06% chance that the percent of all students who played intramural sports who received a degree within six years differs from 57%.
• If the sample proportion of students who played intramural sports who received a degree within six years is 53% and if another 208 students who played intramural sports are surveyed then there would be a 23.06% chance that we would conclude either fewer than 57% of all students who played intramural sports received a degree within six years or more than 57% of all students who played intramural sports received a degree within six years.
• If the population proportion of students who played intramural sports who received a degree within six years is 57% and if another 208 students who played intramural sports are surveyed then there would be a 23.06% chance that either fewer than 53% of the 208 students surveyed students received a degree within six years or more than 61% of the 208 students surveyed received a degree within six years.
• There is a 23.06% chance of a Type I error.

1. Interpret the level of significance in the context of the study.

• If the population proportion of students who played intramural sports who received a degree within six years is 57% and if another 208 students who played intramural sports are surveyed then there would be a 5% chance that we would end up falsely concluding that the proportion of all students who played intramural sports who received a degree within six years is different from 57%
• There is a 5% chance that aliens have secretly taken over the earth and have cleverly disguised themselves as the presidents of each of the countries on earth.
• There is a 5% chance that the proportion of all students who played intramural sports who received a degree within six years is different from 57%.
• If the population proportion of students who played intramural sports who received a degree within six years is different from 57% and if another 208 students who played intramural sports are surveyed then there would be a 5% chance that we would end up falsely concluding that the proportion of all students who played intramural sports who received a degree within six years is equal to 57%.