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?
| Stavka koja se može prevući | arrow_right_alt | Odgovarajuća stavka |
|---|---|---|
True mean weight of pineapples grown in the field this year > 31 oz. | arrow_right_alt | Parameter of interest |
True mean weight of pineapples grown in the field this year. | arrow_right_alt | NOT the parameter of interest |
Mean weight of pineapples | arrow_right_alt | Null Hypothesis |
True mean weight of pineapples grown in the field this year=31 oz. | arrow_right_alt | 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?
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?
| Stavka koja se može prevući | arrow_right_alt | Odgovarajuća stavka |
|---|---|---|
Mean score for the SRS of 45 students at her college | arrow_right_alt | Parameter of interest |
True mean score for college students = 115. | arrow_right_alt | NOT the parameter of interest |
True mean score for college students > 115. | arrow_right_alt | Null Hypothesis |
True mean score for U.S. college students' attitude toward school & study habits | arrow_right_alt | 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?
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?
| Stavka koja se može prevući | arrow_right_alt | Odgovarajuća stavka |
|---|---|---|
Proportion of 150 students that seldom or never argue with their friends. | arrow_right_alt | Parameter of interest |
True proportion does not equal 72% | arrow_right_alt | NOT the parameter of interest |
Proportion = 72% | arrow_right_alt | Null Hypothesis |
True proportion = 72% | arrow_right_alt | Alternative Hypothesis |
True proportion of teens that seldom or never argue with their friends. | arrow_right_alt |
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?