8.1 Practice Writing Hypotheses & Interpreting P Value DUE 5/3

Last updated about 2 years ago

9 questions

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4

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8.1 Practice Writing Hypotheses & Interpreting P Value DUE 5/3

Last updated about 2 years ago

9 questions

Required

4

Draggable item		Corresponding Item

Required

3

In the study of pineapple weights from #1, the sample mean weight from this year’s crop was 31.4 ounces and the sample standard deviation was 2.5 ounces.
A significance test yields a P-value of 0.1317.
Interpret the P-value in context.
I will grade this question.

Required

3

Required

4

Draggable item		Corresponding Item

3

In the study of older students’ attitudes from #4, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.
Interpret the P value in context.
I will grade this question.

Required

3

Required

4

Draggable item		Corresponding Item

Required

3

NEW PROBLEM:
Lumber companies dry freshly cut wood in kilns before selling it. A certain percentage of the boards become “checked,” which means that cracks develop at the ends of the boards during drying. The current drying procedure is known to produce cracks in 16% of the boards.
The drying supervisor at a lumber company wants to test a new method to determine if fewer boards crack. She uses the new method on a random sample of 200 boards and finds that the sample proportion of checked boards is 0.11.

The supervisor performs a test of
Ho: p = 0.16 versus Ha: p < 0.16, where p is the true proportion of all boards that would crack when dried with the new method.

A significance test yields a P-value of 0.027.
Interpret this P-value in context. I will grade this question.

Required

3

At the Hawaii Pineapple Company, managers are interested in the size of the pineapples grown in the company’s fields. Last year, the mean weight of the pineapples harvested from one large field was 31 ounces.
A new irrigation system was installed in this field after the growing season. Managers wonder if the mean weight of pineapples grown in the field this year will be greater than last year. So they take an SRS of 50 pineapples from this year’s crop.
What is the parameter of interest?
What are the null and alternative hypotheses?

Draggable item		Corresponding Item
True mean weight of pineapples grown in the field this year > 31 oz.		Parameter of interest
True mean weight of pineapples grown in the field this year.		NOT the parameter of interest
Mean weight of pineapples		Null Hypothesis
True mean weight of pineapples grown in the field this year=31 oz.		Alternative Hypothesis

In the study of pineapple weights from #1, the sample mean weight from this year’s crop was 31.4 ounces and the sample standard deviation was 2.5 ounces.
A significance test yields a P-value of 0.1317.
Interpret the P-value in context.
I will grade this question.

What conclusion would you make in #2 about the true mean pineapple weight?

Because 13.17% is greater than 5%

we do not have convincing evidence that the mean weight of the pineapples has increased.

we fail to reject the null hypothesis, 

Because 0.1317% is less than 5%

we reject the null hypothesis,

we do have convincing evidence that the mean weight of the pineapples has increased.

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. The mean score for U.S. college students is about 115.
A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 of the more than 1000 students at her college who are at least 30 years of age.
What is the parameter of interest?
What are the null and alternative hypotheses?

Draggable item		Corresponding Item
Mean score for the SRS of 45 students at her college		Parameter of interest
True mean score for college students = 115.		NOT the parameter of interest
True mean score for college students > 115.		Null Hypothesis
True mean score for U.S. college students' attitude toward school & study habits		Alternative Hypothesis

In the study of older students’ attitudes from #4, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.
Interpret the P value in context.
I will grade this question.

What conclusion would you make in #5 about the true mean score for college students' attitudes about school and studying?

we reject the null hypothesis that the mean score = 115,

we do have convincing evidence that the SSHA score is higher than 115 by U.S. college students.

we fail to reject the null hypothesis that the mean score = 115, 

we do not have convincing evidence that the SSHA score is higher than 115 by U.S. college students.

Because 1.01% is < 5% 

Because 10.1% > 5%

A Gallup Poll report revealed that 72% of teens said they seldom or never argue with their friends. Yvonne wonders whether this result holds true in her large high school. She surveys a random sample of 150 students at her school.
What is the parameter of interest?
What are the null and alternative hypotheses?

Draggable item		Corresponding Item
Proportion of 150 students that seldom or never argue with their friends.		Parameter of interest
True proportion does not equal 72%		NOT the parameter of interest
Proportion = 72%		Null Hypothesis
True proportion = 72%		Alternative Hypothesis
True proportion of teens that seldom or never argue with their friends.

NEW PROBLEM:
Lumber companies dry freshly cut wood in kilns before selling it. A certain percentage of the boards become “checked,” which means that cracks develop at the ends of the boards during drying. The current drying procedure is known to produce cracks in 16% of the boards.
The drying supervisor at a lumber company wants to test a new method to determine if fewer boards crack. She uses the new method on a random sample of 200 boards and finds that the sample proportion of checked boards is 0.11.

The supervisor performs a test of
Ho: p = 0.16 versus Ha: p < 0.16, where p is the true proportion of all boards that would crack when dried with the new method.

A significance test yields a P-value of 0.027.
Interpret this P-value in context. I will grade this question.

What conclusion would you make in about the true proportion of all boards that would crack when dried with the new method?

 we have convincing evidence that with the new method a smaller percent of the boards will crack.

Because 27% is greater than 5%

we do not have convincing evidence that with the new method a smaller percent of the boards will crack.

Because 2.7% is less than 5%

we reject the null hypothesis that 16% of the boards will crack with the new method,

we fail to reject the null hypothesis that 16% of the boards will crack with the new method, 

At the Hawaii Pineapple Company, managers are interested in the size of the pineapples grown in the company’s fields. Last year, the mean weight of the pineapples harvested from one large field was 31 ounces. 
A new irrigation system was installed in this field after the growing season. Managers wonder if the mean weight of pineapples grown in the field this year will be greater than last year. So they take an SRS of 50 pineapples from this year’s crop.
What is the parameter of interest?
What are the null and alternative hypotheses?

True mean weight of pineapples grown in the field this year > 31 oz.

Parameter of interest

True mean weight of pineapples grown in the field this year.

NOT the parameter of interest

Mean weight of pineapples

Null Hypothesis

True mean weight of pineapples grown in the field this year=31 oz.

Alternative Hypothesis

What conclusion would you make in #2 about the true mean pineapple weight?

Because 13.17% is greater than 5%

we do not have convincing evidence that the mean weight of the pineapples has increased.

we fail to reject the null hypothesis, 

Because 0.1317% is less than 5%

we reject the null hypothesis,

we do have convincing evidence that the mean weight of the pineapples has increased.

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. The mean score for U.S. college students is about 115. 
A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 of the more than 1000 students at her college who are at least 30 years of age.
What is the parameter of interest?
What are the null and alternative hypotheses?

Mean score for the SRS of 45 students at her college

Parameter of interest

True mean score for college students = 115.

NOT the parameter of interest

True mean score for college students > 115.

Null Hypothesis

True mean score for U.S. college students' attitude toward school & study habits

Alternative Hypothesis

What conclusion would you make in #5 about the true mean score for college students' attitudes about school and studying?

we reject the null hypothesis that the mean score = 115,

we do have convincing evidence that the SSHA score is higher than 115 by U.S. college students.

we fail to reject the null hypothesis that the mean score = 115, 

we do not have convincing evidence that the SSHA score is higher than 115 by U.S. college students.

Because 1.01% is < 5% 

Because 10.1% > 5%

A Gallup Poll report revealed that 72% of teens said they seldom or never argue with their friends. Yvonne wonders whether this result holds true in her large high school. She surveys a random sample of 150 students at her school.
What is the parameter of interest?
What are the null and alternative hypotheses?

Proportion of 150 students that seldom or never argue with their friends.

Parameter of interest

True proportion does not equal 72%

NOT the parameter of interest

Proportion = 72%

Null Hypothesis

True proportion = 72%

Alternative Hypothesis

True proportion of teens that seldom or never argue with their friends.

What conclusion would you make in about the true proportion of all boards that would crack when dried with the new method?

 we have convincing evidence that with the new method a smaller percent of the boards will crack.

Because 27% is greater than 5%

we do not have convincing evidence that with the new method a smaller percent of the boards will crack.

Because 2.7% is less than 5%

we reject the null hypothesis that 16% of the boards will crack with the new method,

we fail to reject the null hypothesis that 16% of the boards will crack with the new method, 

8.1 Practice Writing Hypotheses & Interpreting P Value DUE 5/3

8.1 Practice Writing Hypotheses & Interpreting P Value DUE 5/3

In the study of pineapple weights from #1, the sample mean weight from this year’s crop was 31.4 ounces and the sample standard deviation was 2.5 ounces. A significance test yields a P-value of 0.1317. Interpret the P-value in context.I will grade this question.

In the study of older students’ attitudes from #4, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.Interpret the P value in context.I will grade this question.

In the study of pineapple weights from #1, the sample mean weight from this year’s crop was 31.4 ounces and the sample standard deviation was 2.5 ounces. A significance test yields a P-value of 0.1317. Interpret the P-value in context.I will grade this question.

What conclusion would you make in #2 about the true mean pineapple weight?

In the study of older students’ attitudes from #4, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.Interpret the P value in context.I will grade this question.

What conclusion would you make in #5 about the true mean score for college students' attitudes about school and studying?

What conclusion would you make in about the true proportion of all boards that would crack when dried with the new method?

In the study of pineapple weights from #1, the sample mean weight from this year’s crop was 31.4 ounces and the sample standard deviation was 2.5 ounces.
A significance test yields a P-value of 0.1317.
Interpret the P-value in context.
I will grade this question.

In the study of older students’ attitudes from #4, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.
Interpret the P value in context.
I will grade this question.

In the study of pineapple weights from #1, the sample mean weight from this year’s crop was 31.4 ounces and the sample standard deviation was 2.5 ounces.
A significance test yields a P-value of 0.1317.
Interpret the P-value in context.
I will grade this question.

In the study of older students’ attitudes from #4, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.
Interpret the P value in context.
I will grade this question.