# TV watching.

Item 1
Advertisers fear that users of DVRs (digital video recorders) will “fast forward” past commercials when they watch a recorded program. A leading British pay television company told its advertisers that this effect might be offset because DVR users watch more TV. A sample of 15 DVR users showed a daily mean screen time of 2 hours and 26 minutes with a standard deviation of 14 minutes, compared with a daily mean of 2 hours and 7 minutes with a standard deviation of 12 minutes for a sample of 15 non-DVR users.

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(a-1) Construct a 95 percent confidence interval for the difference in mean TV watching. (Round your answers to 2 decimal places.)

The 95% confidence interval is from to .

(a-2) Would this sample support the company’s claim (i.e., is zero within the confidence interval for the mean difference)?

multiple choice 1
• Yes
• No Correct

(b) State any assumptions that are needed.

multiple choice 2
• The variances are equal. Correct
• The variances are unequal.

2
2points
Item 2
In a market research study, members of a consumer test panel are asked to rate the visual appeal (on a 1 to 10 scale) of the texture of dashboard plastic trim in a mockup of a new fuel cell car. The manufacturer is testing four finish textures. Panelists are assigned randomly to evaluate each texture. The test results are shown below. Each cell shows the average rating by panelists who evaluated each texture. Research question: Is mean rating affected by age group and/or by surface type?

Mean Ratings of Dashboard Surface Texture
Age Group Shiny Satin Pebbled Pattern
Youth (under 21) 6.7 6.6 5.5 4.3
Adult (21 to 39) 5.5 5.3 6.2 5.9
Middle-Age (40 to 61) 4.5 5.1 6.7 5.5
Senior (62 and over) 3.9 4.5 6.1 4.1
________________________________________

(a-1) Which type of ANOVA model should be used?

multiple choice 1
• Two factor, with replication
• Two factor, without replication Correct
• One factor, completely randomized

(a-2) Choose the correct row-effect hypotheses.

a. H0: A1 = A2 = A3 = 0 ⇐⇐
Age group means are the same
H1: Not all the Aj are equal to zero ⇐⇐
Age group means differ
b. H0: A1 ≠ A2 ≠ A3 ≠ 0 ⇐⇐
Age group means are the same
H1: All the Aj are equal to zero ⇐⇐
Age group means differ

multiple choice 2
• a Correct
• b

(a-3) Choose the correct column-effect hypotheses.

a. H0: B1 = B2 = B3 = 0 ⇐⇐
Surface type means are the same
H1: Not all the Bk are equal to zero ⇐⇐
Surface type means differ
b. H0: B1 ≠ B2 ≠ B3 ≠ 0 ⇐⇐
Surface type means are the same
H1: Not all the Bk are equal to zero ⇐⇐
Surface type means differ

multiple choice 3
• a Correct
• b

(b) Fill in the missing data. (Round your SS values to 3 decimal places, F values to 2 decimal places, and other answers to 4 decimal places.)

Table of Means
Mean n Std. Dev
Shiny
Satin
Pebbled
Pattern

Youth (Under 21)
Middle-Age (40 to 61)
Senior (62 and over)
Total
________________________________________

ANOVA Table
Source SS df MS F p-value
Rows (Age Group)
Columns (Surface)
Error
Total
________________________________________

(c) Choose the correct statement.

multiple choice 4
• The texture is significantly affected by age group and by surface type.
• The texture is not significantly affected by age group or by surface type. Correct
• The texture is not significantly affected by surface type, but it is affected by age group.
• The texture is not significantly affected by age group, but it is affected by surface type.
3
2points
Item 3
A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought in to the time repairs are completed. A random sample of 12 repair records showed the following repair times (in days): 9, 5, 6, 7, 8, 5, 6, 6, 5, 5, 6, 5.

(a) H0: μ ≤ 5 days versus H1: μ > 5 days. At α = .05, choose the right option.

a. Reject H0 if tcalc > 1.7960
b. Reject H0 if tcalc < 1.7960

multiple choice 1
• a Correct
• b

(b) Calculate the test statistic. (Round your answer to 3 decimal places.)

Test statistic

(c-1) The null hypothesis should be rejected.

multiple choice 2
• FALSE
• TRUE Correct

(c-2) The average repair time is longer than 5 days.

multiple choice 3
• FALSE
• TRUE Correct

(c-3) At α = .05, is the goal being met?

multiple choice 4
• No Correct
• Yes
4
2points
Item 4
The sodium content of a popular sports drink is listed as 200 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample mean of 211.5 mg with a sample standard deviation of 18.5 mg.

(a) State the hypotheses for a two-tailed test of the claimed sodium content.

a. H0: μ ≥ 200 vs. H1: μ < 200
b. H0: μ ≤ 200 vs. H1: μ > 200
c. H0: μ = 200 vs. H1: μ ≠ 200

multiple choice 1
• a
• b
• c Correct

(b) Calculate the t test statistic to test the manufacturer’s claim. (Round your answer to 4 decimal places.)

Test statistic

(c) At the 5 percent level of significance (α = .05), does the sample contradict the manufacturer’s claim?

Reject CorrectH0. The sample contradicts Correctthe manufacturer’s claim.

(d-1) Use Excel to find the p-value and compare it to the level of significance. (Round your answer to 4 decimal places.)

The p-value is . It is lower Correctthan the significance level of .05.

(d-2) Did you come to the same conclusion as you did in part (c)?

multiple choice 2
• Yes Correct
• No
5
2points
Item 5
A cognitive retraining clinic assists outpatient victims of head injury, anoxia, or other conditions that result in cognitive impairment. Each incoming patient is evaluated to establish an appropriate treatment program and estimated length of stay. To see if the evaluation teams are consistent, 12 randomly chosen patients are separately evaluated by two expert teams (A and B) as shown.

Estimated Length of Stay in Weeks
Patient
Team 1 2 3 4 5 6 7 8 9 10 11 12
A 24 24 52 30 40 30 18 30 18 40 24 12
B 24 20 52 36 36 24 36 16 52 24 24 16
________________________________________

(a) Choose the appropriate hypotheses. Assume μd is the difference in mean lengths of team A and team B.

a. H0: μd ≤ 0 vs. H1: μd > 0
b. H0: μd ≥ 0 vs. H1: μd < 0
c. H0: μd = 0 vs. H1: μd ≠ 0

multiple choice
• a
• b
• c Correct

(b) Specify the decision rule at the .10 level of significance. (Round your answers to 3 decimal places. Negative values should be indicated by a minus sign.)

Reject the null hypothesis if the p-value is less than Correct.10 or if tcalc < or tcalc > .

(c) Find the test statistic tcalc. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.)

tcalc

(d) Find the p-value. (Round your answer to 4 decimal places. A negative value should be indicated by a minus sign.)

p-value

(e) Make a decision.

We do not reject Correctthe null hypothesis.

We can Correctconclude that the evaluator teams are consistent in their estimates.
6
2points
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Item 6
The Environmental Protection Agency (EPA) advocates a maximum arsenic level in water of 10 micrograms per liter. Below are results of EPA tests on randomly chosen wells in a suburban Michigan county. Research question: Is the mean arsenic level affected by well depth and/or age of well?

Arsenic Level in Wells (micrograms per liter)
Age of Well (years)
Well Depth Under 10 10 to 19 20 and Over
Shallow 5.4 6.1 6.8
4.3 4.1 5.4
6.1 5.8 5.7
Medium 3.4 5.1 4.5
3.7 3.7 5.5
4.3 4.4 4.6
Deep 2.4 3.8 3.9
2.9 2.7 2.9
2.7 3.4 4.0
________________________________________

(a-1) Choose the correct row-effect hypotheses.

a. H0: A1 ≠ A2 ≠ A3 ≠ 0 ⇐⇐
well depth means differ
H1: All the Aj are equal to zero ⇐⇐
well depth means are the same
b. H0: A1 = A2 = A3 = 0 ⇐⇐
well depth means are the same
H1: Not all the Aj are equal to zero ⇐⇐
well depth means differ

multiple choice 1
• a
• b

(a-2) Choose the correct column-effect hypotheses.

a. H0: B1 ≠ B2 ≠ B3 ≠ 0 ⇐⇐
age of well means differ
H1: All the Bk are equal to zero ⇐⇐
age of well means are the same
b. H0: B1 = B2 = B3 = 0 ⇐⇐
age of well means are the same
H1: Not all the Bk are equal to zero ⇐⇐
age of well means differ

multiple choice 2
• a
• b

(a-3) Choose the correct interaction-effect hypotheses.

a. H0: Not all the ABjk are equal to zero ⇐⇐
there is an interaction effect
H1: All the ABjk are equal to zero ⇐⇐
there is no interaction effect
b. H0: All the ABjk are equal to zero ⇐⇐
there is no interaction effect
H1: Not all the ABjk are equal to zero ⇐⇐
there is an interaction effect

multiple choice 3
• a
• b

(b) Fill in the missing data. (Round your Table of Means and F values to 2 decimal places, SS values to 3 decimal places, and other answers to 4 decimal places.)

Table of Means
Factor 2: Age of Well
Under 10 10 to 19 20 and Over
Factor 1: Shallow

Depth of well Medium

Deep

Total

________________________________________

ANOVA Table
Source SS df MS F p-value
Factor 1

Factor 2

Interaction

Error

Total

________________________________________

(c) Using α = .05, choose the correct statement.

multiple choice 4
• The main effects of depth of wells and age of wells are significant, but there is not a significant interaction effect.
• The main effect of depth of wells is significant; however, there is no significant effect from age of wells or interaction between Depth of wells and Age of wells.
• The main effect of age of wells is significant; however, there is no significant effect from depth of wells or interaction between Depth of wells and Age of wells.

(d) Calculate the mean for each group and the Tukey test statistic Tcalc for each pair. Provide the critical values for both α = 0.05 and α = 0.01. (Input the mean values within the input boxes of the first row and input boxes of the first column. Input Tcalc in the appropriate boxes in the table. Round all answers to two decimal places.)

Post hoc analysis:

Tukey simultaneous comparison t-values (d.f. = 18)
Deep Medium Shallow

Deep

Medium

Shallow

Critical values for experimentwise error rate:
0.05

0.01

________________________________________

Tukey simultaneous comparison t-values (d.f. = 18)
Under 10 10 to 19 20 and Over

Under 10

10 to 19

20 and Over

Critical values for experimentwise error rate:
0.05

0.01

________________________________________
7
2points
Item 7
The number of entrees purchased in a single order at a Noodles & Company restaurant has had a historical average of 1.5 entrees per order. On a particular Saturday afternoon, a random sample of 42 Noodles orders had a mean number of entrees equal to 1.85 with a standard deviation equal to 0.73. At the 1 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?

(a) Choose the correct null and alternative hypotheses.

a. H0: μ ≥ 1.5 vs. H1: μ < 1.5
b. H0: μ ≤ 1.5 vs. H1: μ > 1.5
c. H0: μ = 1.5 vs. H1: μ ≠ 1.5

multiple choice 1
• a
• b Correct
• c

(b-1) Calculate the t statistic. (Round your answer to 2 decimal places.)

tcalc

p-value

(c) Choose the correct conclusion.

multiple choice 2
• Because the p-value is greater than 0.01, we conclude that there is no evidence to indicate a significant increase in the average number of entrees per order.
• Because the p-value is less than 0.01, we conclude that there is evidence to indicate a significant increase in the average number of entrees per order. Correct
8
2points
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Item 8
Medicare spending per patient in different U.S. metropolitan areas may differ. Based on the sample data below, answer the questions that follow to determine whether the average spending in the northern region is significantly less than the average spending in the southern region at the 1 percent level.

Medicare Spending per Patient (adjusted for age, sex, and race)
Statistic Northern Region Southern Region
Sample mean \$4,346 \$6,795
Sample standard deviation \$1,535 \$3,616
Sample size 12 patients 19 patients
________________________________________

(a-1) Choose the appropriate hypotheses. Assume μN is the average spending in the northern region and μS is the average spending in the southern region.

a. H0: μN − μS ≤ 0 versus H1: μN – μS >0
b. H0: μN − μS ≥ 0 versus H1: μN – μS < 0

multiple choice 1
• a
• b

(a-2) Specify the decision rule. (Use the quick rule to determine degrees of freedom. Round your answer to 3 decimal places. A negative value should be indicated by a minus sign.)

Reject the null hypothesis if tcalc < .

(b) Find the test statistic tcalc assuming unequal variances. (Round your answer to 2 decimal places. A negative value should be indicated by a minus sign.)

tcalc

(c-1) Can we reject the null hypothesis?

multiple choice 2
• No, fail to reject the null hypothesis.
• Yes, reject the null hypothesis.

(c-2) The average spending in the northern region is significantly less than the average spending in the southern region.

multiple choice 3
• False
• True
9
2points
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Item 9
At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 526 ml. The filling process follows a normal distribution with a known process standard deviation of 10 ml.

(a) The normal distribution should be used for the sample mean because (You may select more than one answer. Click the box with a check mark for the correct answer and click to empty the box for the wrong answer.)

check all that apply
• the sample population has a large mean.
• the population distribution is known to be normal.
• the population standard deviation is known.
• the standard deviation is very small.

(b) Set up hypotheses and a two-tailed decision rule for the correct mean using the 5 percent level of significance. The hypothesis for a two-tailed decision is

a. H0: μ ≠526, H1: μ = 526, reject if z < −1.96 or z > 1.96
b. H0: μ ≠526, H1: μ = 526, reject if z > 1.96 or z < −1.96
c. H0: μ = 526, H1: μ ≠ 526, reject if z > 1.96 or z < −1.96
d. H0: μ =526, H1: μ ≠ 526, reject if z > −1.96 or z < 1.96

multiple choice 1
• a
• b
• c
• d

(c) If a sample of 16 bottles shows a mean fill of 532 ml, does this contradict the hypothesis that the true mean is 526 ml?

multiple choice 2
• Yes
• No
10
2points
Item 10
Perfect pitch is the ability to identify musical notes correctly without hearing another note as a reference. The probability that a randomly chosen person has perfect pitch is .0005.

(a) If 36 students at Julliard School of Music are tested, and 1 are found to have perfect pitch, would you conclude at the .01 level of significance that Julliard students are more likely than the general population to have perfect pitch?

multiple choice 1
• No Correct
• Yes

(b) Normality should not be assumed because

a. only 1 students had perfect pitch.
b. the product of n and π0 is too small.
c. all students had perfect pitch.
d. no student had perfect pitch.
multiple choice 2
• a
• b Correct
• c
• d
11
2points
Item 11
The mean arrival rate of flights at O’Hare Airport in marginal weather is 195 flights per hour with a historical standard deviation of 13 flights. To increase arrivals, a new air traffic control procedure is implemented. In the next 30 days of marginal weather, the mean arrival rate is 200 flights per hour.

210 215 200 189 200 213 202 181 197 199
193 209 215 192 179 196 225 199 196 210
199 188 174 176 202 195 195 208 222 221
________________________________________

(a) Set up a right-tailed decision rule at α = .025 to decide whether there has been a significant increase in the mean number of arrivals per hour. Choose the appropriate hypothesis.

a. H1: μ > 195, reject H1 if z < 1.960
b. H1: μ < 195, reject H1 if z > 1.960
c. H0: μ ≥ 195, reject H0 if z < 1.960
d. H0: μ ≤ 195, reject H0 if z > 1.960
multiple choice 1
• a
• b
• c
• d Correct

(b-1) Calculate the test statistic. (Round your answer to 2 decimal places.)

Test statistic

(b-2) What is the conclusion?

multiple choice 2
• There has been a significant increase in the average number of flight departures. Correct
• There has not been a significant increase in the average number of flight departures.

(b-3) Would the decision have been different if you used α = .01?
a. Yes, z.01 = 2.33 and 2.11 < 2.33, so we would have concluded that there is no evidence to indicate a significant increase.
b. No, z.01 = 2.33 and 2.11 < 2.33, so we would have concluded that there has been a significant increase in the average number of flight departures.
multiple choice 3
• Yes Correct
• No

(c) What assumptions are you making, if any?

multiple choice 4
• We have assumed that the type of distribution does not matter.
• We have assumed a normal population or at least one that is not badly skewed. Correct
• We have assumed that the data follow a uniform distribution.
12
2points
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Item 12

Undergraduate Student Rents (n = 10)
820 810 920 690 720
820 800 690 1,020 690
________________________________________

Graduate Student Rents (n = 12)
1,150 950 960 860 750 940
780 810 910 920 900 860
________________________________________

(a) Construct a 90 percent confidence interval for the difference of mean monthly rent paid by undergraduates and graduate students, using the assumption of unequal variances with Welch’s formula for d.f. (Do not round the intermediate calculations. Round your final answers to 3 decimal places. Negative values should be indicated by a minus sign.)

The 90% confidence interval is from to

(b) What do you conclude?

multiple choice
• We can conclude there is a significant difference in means for undergraduate and graduate rent.
• We cannot conclude there is a significant difference in means for undergraduate and graduate rent.
13
2points
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Item 13
The historical on-time percentage for Amtrak’s Sunset Limited is 10 percent. In July, the train was on time 2 times in 30 runs. Hint: Use Excel to calculate the cumulative binomial probability P(X ≤ 2 | n = 30,
ππ
= 0.10). Assume α = 0.1.

(a) Choose the correct null and alternative hypotheses and decision rule.

a. H0:
ππ
≥ .10 versus H1:
ππ
< .10. Reject H0 if P < 0.10
b. H0:
ππ
≥ .10 versus H1:
ππ
< .10. Reject H0 if P > 0.10

multiple choice 1
• a
• b

P(X ≤ 2)

(c-1) Can we reject the null hypothesis?

multiple choice 2
• No
• Yes

(c-2) Has the on-time percentage fallen?

multiple choice 3
• No
• Yes
14
2points
Item 14
The results shown below are mean productivity measurements (average number of assemblies completed per hour) for a random sample of workers at each of three plants. Research question: Are the mean hourly productivity levels the same for workers in these three plants?

Hourly Productivity of Assemblers in Plants
Plant Finished Units Produced per Hour
A (9 workers) 3.6 5.1 2.8 4.6 4.7 4.1 3.4 2.9 4.5
B (6 workers) 2.7 3.1 5.0 1.9 2.2 3.2
C (10 workers) 6.8 2.5 5.4 6.7 4.6 3.9 5.4 4.9 7.1 8.4
________________________________________

Fill in the missing data. (Round your p-value to 4 decimal places, mean values to 2 decimal places, and other answers to 3 decimal places.)

Group Mean n Std. Dev Variance
Plant A
Plant B
Plant C
Total
________________________________________

One-Factor ANOVA
Source SS df MS F p-value
Treatment
Error
Total
________________________________________

Based on the hypotheses given below, choose the correct option.

H0: μ1 = μ2 = μ3
H1: Not all the means are equal

α = .05

multiple choice 1
• Reject the null hypothesis if p-value < 0.05 Correct
• Reject the null hypothesis if p-value > 0.05

We reject the null hypotheses.

multiple choice 2
• No
• Yes Correct

On the basis of the above findings, we conclude that

multiple choice 3
• all the means are below the overall mean.
• all the means are above the overall mean.
• only one mean is above the overall mean.
• one of the means is above and two are below the overall mean. Correct

Calculate the mean for each group and the Tukey test statistic Tcalc for each pair. Provide the critical values for both α = .05 and α = .01.(Input the mean values within the input boxes of the first row and input boxes of the first column. Input Tcalc in the appropriate boxes in the table. Round all answers to two decimal places.)

Tukey simultaneous comparison t-values (d.f. = 22)
Plant B Plant A Plant C

Plant B
Plant A
Plant C
Critical values for experimentwise error rate:
0.05
0.01
________________________________________

Based on the hypothesis given below, choose the correct option.

H0: σ21 = σ22 = σ23
H1: Not all the variances are equal
α = .05

multiple choice 4
• Reject the null hypothesis if H < 6.94
• Reject the null hypothesis if H > 6.94 Correct

Calculate the value of Hcalc. (Round your answer to 2 decimal places.)

Hcalc

On the basis of the above-determined Hcalc, we reject the null hypothesis.

multiple choice 5
• True
• False Correct
15
2points
Item 15
The sodium content of a popular sports drink is listed as 220 mg in a 32-oz bottle. Analysis of 10 bottles indicates a sample mean of 228.2 mg with a sample standard deviation of 18.2 mg.

(a) State the hypotheses for a two-tailed test of the claimed sodium content.
a. H0: μ ≥ 220 vs. H1: μ < 220
b. H0: μ ≤ 220 vs. H1: μ > 220
c. H0: μ = 220 vs. H1: μ ≠ 220
multiple choice 1
• a
• b
• c Correct

(b) Calculate the t test statistic to test the manufacturer’s claim. (Round your answer to 4 decimal places.)

Test statistic

(c) At the 5 percent level of significance (α = .05), does the sample contradict the manufacturer’s claim?

Do not reject CorrectH0. The sample does not contradict Correctthe manufacturer’s claim.

(d-1) Use Excel to find the p-value and compare it to the level of significance. (Round your answer to 4 decimal places.)

The p-value is . It is greater Correctthan the significance level of .05.

(d-2) Did you come to the same conclusion as you did in part (c)?

multiple choice 2
• Yes Correct
• No

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