A Random Sample of 100 Freshmen Students at University Park Campus Is Taken to Estimate

Lesson 10

Question 1

A random sample of 100 freshmen students at University Park campus is taken to estimate the percentage of freshmen that have taken a tour of Pattee Library during orientation week. Of the 100 freshmen, 40 (40%) of them have taken a tour of Pattee Library. The standard error (SE) for a sample percentage in this situation is estimated to be approximately:

4.0%
4.9%
0.24%
24%

Question 2

A random sample of 400 freshmen students at University Park campus is taken to estimate the percentage of freshmen who have taken a tour of Pattee Library during orientation week. Of the 400 freshmen, 160 (40%) of them have taken a tour of Pattee Library. The standard error (SE) for a sample percentage in this situation is estimated to be approximately:

4.9%
4.0%
2.45%
24%

Question 3

When the sample size increases by 4 times (going from a sample of size 100 to a sample of size 400 with everything else remaining the same), the SE of the sampling distribution for our sample proportion is:

½ as large as it was before the sample size increase
2 times larger than it was before the sample size increase
4 times larger than it was before the sample size increase
¼ as large as it was before the sample size increase

Question 4

As sample size increases:

the SE increases
the SE remains the same
the SE decreases

Question 5

A random sample of 400 freshmen students at University Park campus is taken to estimate the percentage of freshmen who have taken a tour of Pattee Library during orientation week. Of the 400 freshmen, 160 (40%) of them have taken a tour of Pattee Library. The 95% confidence interval for the population percentage for all freshmen who have taken a tour of Pattee Library during orientation week is:

40% +/- 2.45%
40% +/- 4.90%
95% +/- 40%
5% +/- 40%

Question 6

An exit poll was done on election night. The exit poll found that 204 out of 400 randomly selected voters (51%) had voted for the incumbent Senator. The 95% confidence interval for the population percentage of all voters who voted for the incumbent Senator in that election is:

51% +/- 5%
51% +/- 2.5%
95% +/- 5%
95% +/- 2.5%

Question 7

A random sample of 225 residents of a small mid-western town found that 80% approve of a new parking ordinance. The 95% confidence interval for the true population percentage of all residents who favor the parking ordinance is:

80% +/- 5.33%
80% +/- 2.666%
80% +/- 8%
80% +/- 2%

Question 8

Based on an exit poll done on election night, the 95% confidence interval for the population percentage of all voters who voted for the incumbent Senator was 52% +/- 3%. Based on this confidence interval, can we state that we are 95% confident that the Senator got a majority of the vote (greater than 50%) and was re-elected?

Yes
No

Question 9

A random sample of 81 residents of a Central Pennsylvania county found that the average restaurant expenditure in the last month was $60 with a sample standard deviation of $18. The standard error (SE) for the average amount spent will be:

$18
$0.22
$2
$6.67

Question 10

A random sample of 100 statistics students finds that the average time spent on an exam is 50 minutes with a sample standard deviation of 5 minutes. The standard error (SE) for the average time spent is:

10.0 minutes
1.0 minutes
5.0 minutes
0.5 minutes

Question 11

A random sample of 400 statistics students finds that the average time spent on an exam is 50 minutes with a sample standard deviation of 5 minutes. The standard error (SE) for the average time spent is:

0.5 minutes
0.25 minutes
0.0125 minutes
5.0 minutes

Question 12

Everything else being equal, an increase in sample size from 100 to 400 will result in the standard error (SE) for the average ______.

decreasing to ¼ of what it was for the sample size of 100
decreasing to ½ of what it was for the sample size of 100
increasing to 2 times what it was for the sample size of 100
increasing to 4 times what it was for the sample size of 100

Question 13

A random sample of 100 full time graduate students in the College of Science is taken to estimate the amount spent on textbooks this semester. The sampled students spent an average of $400 with a sample standard deviation of $100. A 95% confidence interval for the average amount spent by all full time graduate students in the College of Science would be:

$400 +/- $100
$400 +/- $200
$400 +/- $10
$400 +/- $20

Question 14

A random sample of 225 Penn State football season ticket holders is taken to estimate the population average amount spent in State College on a typical home football game weekend. The sample average was $500 with a sample standard deviation of $75. A 95% confidence interval for the average amount spent by all Penn State Football season ticket holders on a typical home football game weekend is:

$500 +/- $75
$500 +/- $5
$500 +/- $10
$500 +/- $150

Question 15

A random sample of 81 Statistics undergraduate students found that the average amount of hours of sleep per week during the past semester was 45.5 hours with a sample standard deviation of 3 hours. A 95% confidence interval for the average amount of hours of sleep per week by all Statistics undergraduate students is:

45.5 +/- 0.666 hours
45.5 +/- 0.333 hours
45.5 +/- 3 hours
45.5 +/- 6 hours

Question 16

A 95% confidence interval for the average amount spent on textbooks by undergraduate students at a large state university per semester is $400 +/- $50. Would it be correct to state that the true population average could be $360?

Yes
No

Question 17

A consumer testing organization is comparing the battery life of two different models of PCs.They randomly sample 49 of each model of PC (independent samples). The 49 PCs in group 1 have an average battery life of 11 hours with a sample standard deviation of 0.7 hours. The 49 PCs in group 2 have an average battery life of 9 hours with a sample standard deviation of 1.4 hours
The standard error for the difference in the two averages is:

0.7 + 1.4 = 2.1 hours
√(0.7^2 + 1.4^2) = 1.56 hours
√(0.1^2 + 0.2^2) = 0.223 hours
0.1 + 0.2 = 0.3 hours

Question 18

A consumer testing organization is comparing the battery life of two different models of PCs. They randomly sample 49 of each model of PC. The 49 PCs in group 1 have an average battery life of 11 hours with a sample standard deviation of 0.7 hours. The 49 PCs in group 2 have an average battery life of 9 hours with a sample standard deviation of 1.4 hours.
The 95% confidence interval of the difference between the two group population means is:

(11-9) +/- 2 * 0.223 = 2 +/- 0.446 hours
(11-9) +/- 1 * 0.223 = 2 +/- 0.223 hours
(11-9) +/- 2 * 1.56 = 2 +/- 3.12 hours
(11-9) +/- 3 * 0.223 = 2 +/- 0.669 hours

Question 19

For large random samples, when we look at sample means or proportions ... What happens to the width of a confidence interval when the confidence level is increased from 68% to 95% (holding all else constant)? Explain briefly.

Question 20

We randomly select 1000 adults from a population of 2 million and also randomly select 1000 adults from a population of 20 million. If the sample standard deviations are the same, how will the margins of error compare for 95% confidence intervals for the true population average? Will the margin of error for the sample from the larger population be greater than, the same, or less than the margin of error for the sample from the smaller population? Explain briefly.

Question 21

We have calculated a 95% confidence interval for the difference between two population proportions. The interval is 4% +/- 5%. Can we conclude that there is a significant difference between the two group population proportions? Yes or No? Explain briefly.

Question 22

A random sample of 900 homeowners in Allegheny County is taken in order to estimate the average amount of money spent by homeowners on landscaping this spring. It turns out that this group spent an average of $200 with a standard deviation of $150. A 95% confidence interval for the average amount spent on landscaping by all Allegheny County homeowners is calculated to be $200 ± $10. Which is true about this situation?

The true average amount spent must be in this interval.
The procedure used here produces intervals that cover the right answer for 95% of all samples.
This confidence interval won’t cover the right answer since a histogram of the amount spent is skewed.
The interval would be wider if we used an 80% confidence level instead of 95%.

Question 23

A random sample of 900 homeowners in Allegheny County is taken in order to estimate the average amount of money spent by homeowners on landscaping this spring. It turns out that this group spent an average of $200 with a standard deviation of $150. In this situation, what is the population? What is the parameter? What is the sample? What is the statistic?

Question 24

One thousand students are randomly selected from the list of those currently registered at Harvard University, and they each report how many miles they rode in an MTA bus during the past week. Because many students did not ride a bus at all, a histogram of the values does not look like the normal curve. True or False: Even though the histogram did not follow the normal curve, it is still possible to use the normal curve to make a confidence interval for the average number of miles that all Harvard students rode on MTA buses last week. (explain briefly why you chose your answer).

Use for Questions 1-6: A random sample of 100 males found that 80% watched a particular TV show in the past week. A random sample of 100 females found that 50% watched the same TV show in the past week

Question 1

The standard error for the sampling distribution of the sample proportion for males is:

0.04
0.0016
0.08
0.16

Question 2

The standard error for the sampling distribution of the sample proportion for females is:

0.05
0.25
0.0025
0.10

Question 3

What is the standard error for the difference between our two sample values?

√((0.04)2 + (0.05)2) = √0.0041 = 0.064
√(0.04 + .05) = √0.09 = 0.30
√((0.0016)2 + (0.0025)2) = √0.000009 = 0.003
√(0.08 + 0.10) = √0.18 = .42

Question 4

What is the formula for a 95% confidence interval of the difference between the two population proportions?

(sample proportion 1 - sample proportion 2) +/- 2 *SED
(sample proportion 1 - sample proportion 2) +/- 1*SED
(sample proportion 1 - sample proportion 2) +/- 3*SED
(sample proportion 1 - sample proportion 2) +/- 2

Question 5

What is the 95% confidence interval of the difference between the two population proportions?

(.80 - .50) +/- 2 * 0.064 = .30 +/- 0.128 = (0.172 to 0.428) = (17.2% to 42.8%)
(.80 - .50) +/- 1 * 0.003 = .30 +/- 0.003 = (0.297 to 0.303) = (29.7% to 30.3%)
(.80 - .50) +/- 2 * 0.003 = .30 +/- 0.006 = (0.294 to 0.306) = (29.4% to 30.6%)
(.80 - .50) +/- 3 * 0.003 = .30 +/- 0.009 = (0.291 to 0.309) = (29.1% to 30.9%)
(.80 - .50) +/- 1 * 0.064 = .30 +/- 0.064 = (0.236 to 0.364) = (23.6% to 36.4%)
(.80 - .50) +/- 3 *0.064 = .30 +/- 0.192 = (0.108 to 0.492) = (10.8% to 49.2%)

Question 6

The confidence interval in #5 shows that:

the difference between the population proportions for the two groups is significant
the difference between the population proportions for the two groups is not significant
cannot tell from the information given

Question 7

A random sample of 100 full time students at a large two-year college is taken to estimate the amount they spent on textbooks this semester. The sampled students spent an average of $300 with a standard deviation of $50. Thus, a 95% confidence interval for the average amount spent by all full time students at that school would be:

$300 +/- $0.50
$300 +/- $5.00
$300 +/- $10.00
$300 +/- $50.00

Question 8

A random sample of 225 male students at a large two-year college is taken to estimate the height of male students at this school. The sampled male students have an average height of 68 inches with a standard deviation of 9 inches. A random sample of 144 female students is also taken from the same school to estimate the height of female students at this school. The sampled female students have an average height of 64 inches with a standard deviation of 3 inches. A 95% confidence interval for the difference in average height between males and females is given by:

4 +/- 0.92195 inches
4 +/- 1.30
4 +/- 1.844 inches
4 +/- 0.65 inches

Lesson 11

Question 1

A polling group surveyed a city in Scotland regarding residents’ opinions on independence from the UK. The poll wants to find out whether there is strong evidence to show that a majority (> 50%) of the residents will vote ‘Yes.’ The survey polled 2000 residents, of which 1050 responded that they will vote ‘Yes’ on Scotland independence (52.5%). What are the null and alternative hypotheses?

Null: the percentage of ‘Yes’ votes is 52.5%; Alternative: the percentage of ‘Yes’ votes is greater than 52.5%
Null: the percentage of ‘Yes’ votes is greater than 52.5%; Alternative: the percentage of ‘Yes’ votes is 52.5%
Null: the percentage of ‘Yes’ votes is 50%; Alternative: the percentage of ‘Yes’ votes is greater than 50%
Null: the percentage of ‘Yes’ votes is greater than 50%; Alternative: the percentage of ‘Yes' votes is 50%

Question 2

For patients with a particular disease, the population proportion of those successfully treated with a standard treatment that has been used for many years is 0.75. A medical research group invents a new treatment that they believe will be more successful, i.e. the population proportion will exceed 0.75. A doctor plans a clinical trial he hopes will prove this claim. A sample of 100 patients with the disease is obtained. Each person is treated with the new treatment and eventually classified as having either been successfully or not successfully treated with the new treatment. Out of 100 patients, 80 (80%) were successfully treated by the new treatment. What are the null and alternative hypotheses?

Null: the population proportion of those successfully treated by the new treatment exceeds 0.75 (p > 0.75); Alternative: the population proportion of those successfully treated by the new treatment is 0.75 (p = 0.75)
Null: the population proportion of those successfully treated by the new treatment is 0.75 (p = 0.75); Alternative: the population proportion of those successfully treated by the new treatment exceeds 0.75 (p > 0.75)
Null: the population proportion of those successfully treated by the new treatment is 0.80 (p = 0.80); Alternative: the population proportion of those successfully treated by the new treatment exceeds 0.80 (p > 0.80)
Null: the population proportion of those successfully treated by the new treatment exceeds 0.80 (p > 0.80); Alternative: the population proportion of those successfully treated by the new treatment is 0.80 (p = 0.80)

Question 3

Suppose that a study is done comparing two different contact lens wetting solutions with regard to hours of wearing comfort. 100 contact lens wearers are randomly divided into two groups. One group uses solution A for 2 months. The other group uses solution B for 2 months. The researcher wants to determine if there is a difference in the hours of wearing comfort for the two groups. The population mean number of hours of wearing comfort will be compared for the two groups. What are the null and alternative hypotheses being tested by the researcher?

Null: there is a difference in the population mean number of hours of wearing comfort for the two groups (two population means are not equal); Alternative: there is no difference in the population mean number of hours of wearing comfort for the two groups (two population means are equal).
Null: there is no difference in the population mean number of hours of wearing comfort for the two groups (two population means are equal); Alternative: the population mean from group A is larger than that of group B (population mean group A > population mean group B)
Null: there is no difference in the population mean number of hours of wearing comfort for the two groups (two population means are equal); Alternative: there is a difference in the population mean number of hours of wearing comfort for the two groups (two population means are not equal)
Null: the population mean number of hours of wearing comfort for group A is less than that of group B (population mean A population mean A)

Question 4

A car company is testing to see if the proportion of all adults who prefer blue cars has changed (differs) from 0.35 since industry statistics indicate this proportion has been 0.35 for quite some time. A random sample of 1,000 car owners finds that the proportion that prefers blue cars is 0.40. What are the null and alternative hypotheses being tested?

Null: the population proportion who prefer blue cars is 0.35; Alternative: the population proportion who prefer blue cars is greater than 0.35
Null: the population proportion who prefer blue cars is 0.40; Alternative: the population proportion who prefer blue cars differs from (does not equal) 0.40
Null: the population proportion who prefer blue cars is 0.40; Alternative: the population proportion who prefer blue cars is greater than 0.40
Null: the population proportion who prefer blue cars is 0.35; Alternative: the population proportion who prefer blue cars differs from (does not equal) 0.35

Question 5

A college admissions website tells potential students that the average amount spent on textbooks is $300.00 per semester. A student group decides to research book cost because it now believes that book cost per semester is greater than $300.00. A random sample of 225 current students finds that the sample average is $324 with a standard deviation of $60. In this situation, the null hypothesis is that:

the average for all students is greater than $300
the average for all students is greater than $324
the average for all students equals $300
the average for all students equals $324

Question 6

$60
$4
$8
$15

Question 7

+24
+6
-6
0.4

Question 8

A polling organization surveyed Pennsylvanians regarding residents’ opinions on whether there should be term limits for State Representatives and Senators. The polling organization wants to find out if there is strong evidence that a majority (> 50%) of the residents are in favor of term limits. The pollsters randomly polled 500 residents, of which 275 responded that they favor term limits (55%). What approximately is the standard error of the proportion (SEP) under the null hypothesis?

0.044 or 4.4%
0.011 or 1.1%
0.022 or 2.2%
0.05 or 5%

Question 9

0.40
0.05
-3.33
+3.33

Question 10

A consultant believes that the average amount spent by customers at an online shoe store will be less than the current $100 if shipping costs are increased by 10% (the projected shipping cost increase proposed by their carrier). To test the null hypothesis that the population mean = $100 versus the alternative hypothesis that the population mean < $100, the consultant conducts a study using a large random sample and calculates a test statistic of z = +0.5. The p-value for this test would be