Question 4(1 point)

Question 4 options:

The makers of a smartphone have received complaints that the face recognition tool often doesn't work, or takes multiple attempts to finally unlock the phone. They've upgraded to a new version and are claiming the tool has improved. To test the claim, a critic takes a random sample of 80 users of the new version (Group 1) and 75 users of the old version (Group 2). He finds that the face recognition tool works on the first try 75% of the time in the new version and 60% of the time in the old version. Can it be concluded that the new version is performing better? Test at =0.10.

Hypotheses:

H₀:p₁=p₂

H₁:p₁>p₂

In this scenario, what is the test statistic? Round to four decimal places.

z =_____

Question 5(1 point)

Question 5 options:

A math teacher tells her students that eating a healthy breakfast on a test day will help their brain function and perform well on their test. During finals week, she randomly samples 46 students and asks them at the door what they ate for breakfast. She categorizes 26 students into Group 1 as those who ate a healthy breakfast that morning and 20 students into Group 2 as those who did not. After grading the final, she finds that 50% of the students in Group 1 earned an 80% or higher on the test, and 40% of the students in Group 2 earned an 80% or higher. Can it be concluded that eating a healthy breakfast improves test scores? Use a 0.05 level of significance.

Hypotheses:

H₀:p₁=p₂

H₁:p₁>p₂

In this scenario, what is the test statistic? (Round to 4 decimal places)

z =

Question 6(1 point)

Which of the following statements are true of the null and alternative hypotheses?

Question 6 options:

Both hypotheses must be true

It is possible for both hypotheses to be true

It is possible for neither hypothesis to be true

Exactly one hypothesis must be true

Question 7(1 point)

A lab technician is tested for her consistency by taking multiple measurements of cholesterol levels from the same blood sample. The target accuracy has an average of 2.78 or less with a standard deviation of 1.17. If the lab technician takes 16 measurements and the variance of the measurements in the sample is 2.45, does this provide enough evidence to reject the claim that the lab technician's accuracy is within the target accuracy?

Compute the value of the appropriate test statistic.

Question 7 options:

t = -1.345

t = -1.645

t = -1.128

t = -1.28

Question 8(1 point)

In an article appearing inToday's Healtha writer states that the average number of calories in a serving of popcorn is 76. To determine if the average number of calories in a serving of popcorn is different from 76, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 80 with a sample standard deviation of 6.3.

State the null and alternative hypotheses.

Question 8 options:

In an article appearing inToday's Healtha writer states that the average number of calories in a serving of popcorn is 76. To determine if the average number of calories in a serving of popcorn is different from 76, a nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean number of calories per serving to be 80 with a sample standard deviation of 6.3.

State the null and alternative hypotheses.

Question 8 options:

Question 9(1 point)

A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. Suppose that a storm is in progress with a severe storm class rating. Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down. The next 10 waves have an average wave height is 16 feet with a SD of .6 ft.

If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? What is the p-value?

Question 9 options:

H₁:>16.4 feet; the P-value = 0.0321

H₁:< 16.4 feet; the P-value =0.0321

H₁:> 16.4 feet; the P-value =.9757

H₁:=16.4 feet; the P-value =.9757

Question 10(1 point)

The plant-breeding department at a major university developed a new hybrid boysenberry plant called Stumptown Berry. Based on research data, the claim is made that from the time shoots are planted 90 days on average are required to obtain the first berry. A corporation that is interested in marketing the product tests 60 shoots by planting them and recording the number of days before each plant produces its first berry. The sample mean is 92.3 days and the SD is 6.66 days. The corporation wants to know if the mean number of days is different from the 90 days claimed? Use alpha = .01.

Question 10 options:

No, because the p-value =.0048

Yes, because the p-value = .0097

No, because the p-value = .0097

No, because the p-value = .9873

Yes, because the p-value = .0048

Question 11(1 point)

A movie theater company wants to see if there is a difference in the average movie ticket sales in San Diego and Portland per week. They sample 20 sales from San Diego and 20 sales from Portland over a week. Test the claim using a 5% level of significance. Assume the variances are unequal and that movie sales are normally distributed.

San Diego

Portland

234

211

221

214

202

228

214

222

228

218

244

216

182

222

245

220

215

228

233

224

227

234

217

219

219

226

234

226

255

219

235

228

211

212

248

216

232

217

233

214

Choose the correct decision and summary based on the p-value.

Question 11 options:

Do not reject H_0.There is evidence that the average movie ticket sales in San Diego and Portland per week differ.

Reject H₀. There is no evidence that the average movie ticket sales in San Diego and Portland per week differ.

Reject H₀. There is evidence that the average movie ticket sales in San Diego and Portland per week differ.

Do not reject H₀. There is no evidence that the average movie ticket sales in San Diego and Portland per week differ.

Question 12(1 point)

You are testing the claim that the mean GPA of night students is different than the mean GPA of day students. You sample 30 night students and 30 day students. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal.

t-Test: Two-Sample Assuming Unequal Variances

Night

Day

Mean

2.98767

3.3432

Variance

0.29009

0.266482

Observations

30

30

Hypothesized Mean Difference

0

df

58

t Stat

-2.6102

P(T<=t) one-tail

0.0057

t Critical one-tail

1.6716

P(T<=t) two-tail

0.01149

t Critical two-tail

2.0017

The Hypotheses for this problem are:

H₀: ₁= ₂

H₁: ₁₂

Choose the correct decision and summary based on the p-value.

Question 12 options:

Reject H₀. There is not enough evidence that the mean GPA of night students is different than the mean GPA of day students.

Do not reject H_0.There is significant evidence that the mean GPA of night students is different than the mean GPA of day students.

Do not reject H₀. There is not enough evidence that the mean GPA of night students is different than the mean GPA of day students.

Reject H₀. There is significant evidence that the mean GPA of night students is different than the mean GPA of day students.

Question 13(1 point)

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. "Wrinkle recovery angle" measures how well a fabric recovers from wrinkles. Higher is better. Here are data on the wrinkle recovery angle (in degrees) for a random sample of fabric specimens. Assume the populations are approximately normally distributed with unequal variances. A manufacturer believes that the mean wrinkle recovery angle for Hylite is better. A random sample of 25 Hylite (group 1) and 20 Permafresh (group 2) were measured. Test the claim using a 10% level of significance.

Hylite

Permafresh

141

124

140

145

138

136

138

117

138

133

141

139

141

105

132

127

141

137

135

139

146

165

142

131

139

137

131

147

133

148

146

142

139

116

146

148

137

133

142

136

137

134

137

138

143

The Hypotheses for this problem are:

H₀: ₁= ₂

H₁: ₁> ₂

Find the p-value. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal.

p-value=___

___

Question 13 options:

Question 14(1 point)

A professor wants to know whether or not there is a difference in comprehension of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations. She randomly divides her class into two groups of 15, gives one group instructions in text and the second group instructions with visual illustrations. The following data summarizes the scores the students received on a test given after the lab. Let the populations be normally distributed with a population standard deviation of 5.32 points for both the text and visual illustrations.

Text (Group 1)

Visual Illustrations (Group 2)

57.3

59.15

45.3

57.6

87.1

72.9

65.2

83.2

43.1

64

87.3

77.7

75.2

78.2

88.2

64.4

67.5

89

86.2

72.9

68.2

88.2

54.4

43.8

93

98.1

89.2

95.1

55

85.1

Is there evidence to suggest that a difference exists in the comprehension of the lab based on the test scores? Use=0.05.

Enter theP-Value - round to 4 decimal places. Make sure you put the 0 in front of the decimal.

p-value =___

___

Question 14 options:

Question 15(1 point)

A liberal arts college in New Hampshire implemented an online homework system for their introductory math courses and wanted to know whether or not the system improved test scores. In the Fall semester, homework was completed the old fashioned way - with pencil and paper, checking answers in the back of the book. In the Spring semester, homework was completed online - giving students instant feedback on their work. The results are summarized below. Population standard deviations were used from past studies.

Online (Group 1)

Pen and Paper (Group 2)

Number of Students

144

127

Mean Test Score

78.4

75

Population Standard Dev. Test Score

11.98

10.56

Is there evidence to suggest that the online system improves test scores? Use =0.05.

Select the correct alternative hypothesis and state the p-value.

Question 15 options:

H1: 1 2;p-value = .0130

H1: 1 < 2;p-value = .0130

H1: 1 > 2;p-value = .0065

H1: 1 < 2;p-value = .0065

H1: 1 2;p-value = .0065

H1: 1 > 2;p-value = .0130

Question 16(1 point)

The CEO of a large manufacturing company is curious if there is a difference in productivity level of her warehouse employees based on the region of the country the warehouse is located. She randomly selects 35 employees who work in warehouses on the East Coast (Group 1) and 35 employees who work in warehouses in the Midwest (Group 2) and records the number of parts shipped out from each for a week. She finds that East Coast group ships an average of 1287 parts and knows the population standard deviation to be 348. The Midwest group ships an average of 1449 parts and knows the population standard deviation to be 298.

Using a 0.01 level of significance, test if there is a difference in productivity level. What are the correct hypotheses for this problem?

Question 16 options:

H0: 1= 2 ; H1: 1 2

H0: 1 2 ; H1: 1 2

H0: 1 2 ; H1: 1 = 2

H0: 1 2 ; H1: 1 2

H0: 1 = 2 ; H1: 1 = 2

H0: 1 > 2 ; H1: 1 = 2

Question 17(1 point)

You are running a left-tailed 2-sample z-test at the 0.05 level of significance and you find a test statistic of z=-1.87. Shoudl you reject or not reject the null?

Question 17 options:

Reject the null.

Do not reject the null.

Question 18(1 point)

Match the following symbol with the correct phrase.

Question 18 options:

significance level

confidence level

parameter

power

P(Type II Error)

Question 19(1 point)

Match the following symbol with the correct phrase.

Question 19 options:

significance level

confidence level

parameter

power

P(Type II Error)

Question 20(1 point)

The plant-breeding department at a major university developed a new hybrid boysenberry plant called Stumptown Berry. Based on research data, the claim is made that from the time shoots are planted 90 days on average are required to obtain the first berry. A corporation that is interested in marketing the product tests 60 shoots by planting them and recording the number of days before each plant produces its first berry. The corporation wants to know if the mean number of days is different from the 90 days claimed.

A random sample was taken and the following test statistic was z = -2.15 and critical values of z = 1.96 was found.

What is the correct decision and summary?

Question 20 options:

RejectH₀, there is enough evidence to support the corporation's claim that the mean number of days until a berry is produced is different from 90 days claimed by the university.

RejectH₁, there is not enough evidence to reject the corporation's claim that the mean number of days until a berry is produced is different from 90 days claimed by the university.

AcceptH₀, there is enough evidence to support the corporation's claim that the mean number of days until a berry is produced is different from 90 days claimed by the university.

RejectH₀, there is not enough evidence to support the corporation's claim that the mean number of days until a berry is produced is different from 90 days claimed by the university.

Do not rejectH₀, there is not enough evidence to support the corporation's claim that the mean number of days until a berry is produced is different from 90 days claimed by the university.