Statistics- Hinton Motors is testing an experimental battery-powered

1. Hinton Motors is
testing an experimental battery-powered engine for its passenger cars. Hinton
has tested the range of the engine in five different trials, recording the
distance traveled in each trial before the battery needed recharging. Trial
results are given in the table:

Trial	Distance (in miles)
1	840
2	820
3	790
4	850
5	700

Produce the 90%
confidence interval of the average distance for the population represented
here. Assume that the population distribution is approximately normal. Report the
upper bound for your interval.

	831.46
	768.92

	799.36
	857.61

2.
Plainfield Telemarketing plans to estimate the average number of
telephone contacts made by its 245 sales reps over the past month. A sample of
reps will be randomly selected for the estimate. To construct a 99% confidence
interval estimate in which the margin of error will be no more than 20
contacts, how many reps should be included in the sample? Assume a similar
study showed a standard deviation of 120 contacts.

	268
	145

	169
	121

240

3.
In a random sample of 5,000 recent government contracts, 19.3 percent of
the contracts involved cost overruns. Build a 99% confidence interval estimate
of all recent government contracts that involve cost overruns. Report the upper
bound for your interval.

	.087
	.235

	.199
	.146

4.

You want to estimate the proportion of households in the Portland area
that have no land-line telephone service. Suppose you plan to take a simple
random sample of Portland households and target a margin of error no larger
than plus or minus .02 for a proposed 90% confidence interval estimate. How
large a sample would you recommend? (Assume you have no prior

information about the likely population proportion.)

	1026
	1702

	3425
	1322

2963

5.
Estimates vary widely as to the prevalence of cheating on campus. You
plan to administer an anonymous survey to estimate the proportion of students
who have cheated on a test in the past year. At a 95% confidence level, you
would like your estimate to have a margin of error no greater than 3%. How large
a sample should you take to ensure that you achieve your target margin of
error?

	267 students
	757 students

	1068 students
	Insufficient information is given to determine the necessary sample size.

6. Greg would like to
test whether his running pace differs between the morning and the afternoon. In
a sample of 10 morning runs, his average pace was 7 minutes 50 seconds per mile
(490 seconds/mile), with a sample standard deviation of 35 seconds per mile. In
a sample of 12 afternoon runs, his average pace was 7 minutes, 30 seconds per
mile (450 seconds/mile), with a sample standard deviation of 32 seconds per
mile.

Use the information
above to construct a 95% confidence interval for the difference between Greg’s
morning and afternoon pace. Assume that the population standard deviations are
equal. Complete the following sentence: “I am 95% confident that Greg’s
afternoon pace is between __________________________________ faster than his
morning pace.”

	10.2 seconds/mile and 69.8 seconds/mile
	12.0 seconds/mile and 68.0 seconds/mile

	15.3 seconds/mile and 64.7 seconds/mile
	34.8 seconds/mile and 45.2 seconds/mile

7.
A Seattle
Times poll of 503 voters found that 30% of those surveyed favored the incumbent
mayor in Seattle’s upcoming mayoral election. Using a 95% confidence level, the
margin of error for the poll result was reported to be 4.5%. Choose the best
interpretation of this poll result.

	The share of the sample that favors the incumbent mayor is between 25.5% and 34.5%, with 95% confidence.
	The share of voters that favors the incumbent mayor is between 25.5% and 34.5%, with 95% confidence.

	There is a 95% chance that support for the mayor will be between 25.5% and 34.5% in the upcoming election.
	There is a 95% chance that 30% of voters currently favor the mayor in the upcoming election.

8.
Bargain.com
claims to have an average of 637 visitors to its site per hour. You take a
random sample of 100 hours and find the average number of visitors for the
sample is 632 per hour. Can we use this sample result to reject a µ≥ 637 visitors null hypothesis at the 5%
significance level? Assume the population standard deviation is 18 visitors

	Yes. Since the sample mean of 632 is less than the test’s critical value of 634, we can reject the null hypothesis.
	No. Since the sample mean of 632 is greater than the test’s critical value of 628, we can’t reject the null hypothesis.

	Yes. Since the sample mean of 632 is less than the test’s critical value of 639, we can reject the null hypothesis.
	No. Since the sample mean of 632 is greater than the test’s critical value of 614, we can’t reject the null hypothesis.

9. The competing hypotheses for a hypothesis test are as
follows:

Ho: µ≤ 1000
(The proportion mean is less than or equal to 1000)

Ha: µ > 1000 (The
population mean is greater than 1000)

Assume the population
standard deviation is known to be 80. A random sample of size 64 has a sample
mean of 1020. Use the p-value approach, and a significance level of 5%, to decide
whether you can reject the null hypothesis.

	The p-value for this sample result is .0228. Since this p-value is less than the significance level of .05, we can’t reject the null hypothesis.
	the p-value for this sample result is .0428. Since this p-value is less than the significance level of .05, we can’t reject the null hypothesis.

	The p-value for this sample result is .0428. Since this p-vlaue is less than the significance level of .05, we can reject the null hypothesis.
	The p-value for this sample result is .0228. Since this p-value is less than the significance level of .05, we can reject the null hypothesis.

10. Suppose you are testing the following hypotheses:

Ho: µ = 650

Ha: µ≠ 650

Sample size is 100. The
sample mean is 635. The population standard deviation is 140. The significance
level is .10. Compute the sample test statistic, z, and use it to decide
whether to reject the null hypothesis.

	z = -1.87. Since this value is outside the interval -1.65 to +1.65, we can reject the null hypothesis.
	z = -2.17. Since this value is outside the interval -1.65 and +1.65, we can reject the null hypothesis.

	z = -1.37. Since this value is between -1.65 and +1.65, we can’t reject the null hypothesis.
	z = -1.07. Since this value is between -1.65 and +1.65, we can’t reject the null hypothesis.

11. Trend Home Products wants its online service
agents to respond to posted customer complaints in an average of 28 minutes.
You track a random sample of 12 complaints. The average response time for the
sample is 30.2 minutes. The sample standard deviation is 14.3 minutes. Is this
sample result sufficient to reject a null hypothesis that the average response
time for all customer complaints is exactly 28 minutes? Use a 5% significance
level. Assume the population distribution is approximately normal.

	t = .53. Since t is between -2.201 to +2.201, we can’t reject the null hypothesis.
	t = 2.73. Since t is outside the interval -2.332 to + 2.332, we can reject the null hypothesis.

12.

	t = 2.73. Since t is outside the interval -2.201 to +2.201, we can reject the null hypothesis.
	t = 1.33. Since t is between -2.201 to +2.201, we can’t reject the null hypothesis.

13. BarnabyToys.com claims
that the average delivery time for items purchased on its website is 3.65 days.
A random sample of 45 recent deliveries has an average delivery time for the
sample of 4.03 days.

Compute the p-value
for the sample result and use it to test the null hypothesis that average
delivery time for the population of BarnabyToys.com items purchased online is
not more than 3.65 days. Use a 5% significance level for the test and assume
that the population standard deviation is .92 days. is the sample result
sufficient to reject the null hypothesis?

	Since the p-value of .0328 is less than the significance level of .05, we can reject the null hypothesis.
	Since the p-value of .0028 is less than the significance level of 05, we can reject the null hypothesis.

14.

	Since the p-value of .0328 is less than the significance level of .05, we can’t reject the null hypothesis.
	Since the p-value of .0028 is less than the significance level of .05, we can’t reject the null hypothesis.

13 . At a local coffee shop
last year, the average amount spent per customer (APC) was $5.24. In order to
determine whether there has been a change in the APC, the coffee shop took a
sample of 500 recent transactions. For the 500 transactions, the APC was $5.39
with a sample standard deviation of $1.20. The median transaction was $5.15.

The hypotheses test
was constructed as follows:

Ho: µ = 5.24

Ha: µ≠ 5.24

At a 5% significance
level, which of the following is the correct conclusion?

	Reject the null hypothesis because the test statistic of 2.80 is not between -1.65 and 1.65. There is not sufficient sample evidence of a change in the average amount of the transaction.
	Reject the null hypothesis because the test statistic of 2.80 is not between -1.96 and 1.96. There is not sufficient evidence of a change in the average transaction amount.

	Reject the null hypothesis because the test statistic of 2.80 is not between -1.65 and 1.65. There is sufficient sample evidence to argue that the average transaction amount has changed since last year.
	Reject the null hypothesis because the test statistic for 2.80 is not between -1.96 and 1.96. There is sufficient sample evidence to argue that the average transaction amount has changed since last year.