Assignment 6

Ho: p = 0.01 Ha: p < 0.01

A colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. A new study provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01? (a) What are the null and alternative hypotheses?

p̂ = .0046

A colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. A new study provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01? (b) What is the sample proportion? Round your answer to four decimal places

p-value equals .004

A colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. A new study provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01? (c) The figure below shows a randomization distribution for this test. Use the fact that there are 1000 dots in the distribution to find the p-value. (See: https://imgur.com/gWwCyMh)

Yes

A colonoscopy is a screening test for colon cancer, recommended as a routine test for adults over age 50. A new study provides the best evidence yet that this test saves lives. The proportion of people with colon polyps expected to die from colon cancer is 0.01. A sample of 2602 people who had polyps removed during a colonoscopy were followed for 20 years, and 12 of them died from colon cancer. Does this provide evidence that the proportion of people who die from colon cancer after having polyps removed in a colonoscopy is significantly less than the expected proportion (without a colonoscopy) of 0.01? (d) Does the p-value appear to show significant evidence that colonoscopies save lives?

0.07

A randomization distribution is given along with the relevant null and alternative hypotheses. Which p-value is the best estimate for the p-value corresponding to the given observed statistic? (See: https://imgur.com/sfo26S9) Ho: μ = 12 Ha: μ < 12 x̅ = 8

.023

An experiment is conducted to study the effects of smiling on leniency in judging students accused of cheating. The full data is not given in this question. We consider hypotheses Ho: μs = μn vs Ha: μs > μn to test if the data provide evidence that average leniency score is higher for smiling students (μs) than for students with a neutral expression (μn). A dotplot for the difference in sample means based on 1000 random assignments of leniency scores from the original sample to smile and neutral groups is shown below: (See: https://imgur.com/jMWBF68) (a) The difference in sample means for the original sample is D: x̅s - x̅n = 4.91 - 4.12 = 0.79 (as shown above). What is the p-value for the one-tailed test? Enter the exact answer.

No, Yes

An experiment is conducted to study the effects of smiling on leniency in judging students accused of cheating. The full data is not given in this question. We consider hypotheses Ho: μs = μn vs Ha: μs > μn to test if the data provide evidence that average leniency score is higher for smiling students (μs) than for students with a neutral expression (μn). A dotplot for the difference in sample means based on 1000 random assignments of leniency scores from the original sample to smile and neutral groups is shown below: (See: https://imgur.com/jMWBF68) (b) Now we consider the test with a two-tailed alternative, Ho: μs = μn vs Ha: μs ≠ μn, where we make no assumption in advance whether smiling helps or discourages leniency. Does the randomization distribution change for this test? Does the p-value change?

p-value is 0.0029

Arsenic-based additives in chicken feed have been banned by the European Union, but are mixed in the diet of about 70 % of the 9 billion broiler chickens produced annually in the US. Many restaurant and supermarket chains are working to reduce the amount of arsenic in the chicken they sell. To accomplish this, one chain plans to measure, for each supplier, the amount of arsenic in a random sample of chickens. The chain will cancel their relationship with a supplier if the sample provides sufficient evidence that the average amount of arsenic in chicken provided by that supplier is greater than 80 ppb (parts per billion). The hypotheses are: Ho: μ = 80 Ha: μ > 80 where mu represents the mean arsenic level in all chicken meat from that supplier. Samples from two different suppliers are analyzed, and the resulting p-value are given. Sample from Supplier A: p-value is 0.21 Sample from Supplier B: p-value is 0.0029 (a) Which p-value shows stronger evidence for the alternative hypothesis?

Supplier A

Arsenic-based additives in chicken feed have been banned by the European Union, but are mixed in the diet of about 70 % of the 9 billion broiler chickens produced annually in the US. Many restaurant and supermarket chains are working to reduce the amount of arsenic in the chicken they sell. To accomplish this, one chain plans to measure, for each supplier, the amount of arsenic in a random sample of chickens. The chain will cancel their relationship with a supplier if the sample provides sufficient evidence that the average amount of arsenic in chicken provided by that supplier is greater than 80 ppb (parts per billion). The hypotheses are: Ho: μ = 80 Ha: μ > 80 where mu represents the mean arsenic level in all chicken meat from that supplier. Samples from two different suppliers are analyzed, and the resulting p-value are given. Sample from Supplier A: p-value is 0.21 Sample from Supplier B: p-value is 0.0029 (b) Which supplier, A or B, should the chain get chickens from in order to avoid too high a level of arsenic?

No conclusive evidence of anything

Hypotheses for a statistical test are given. Ho: p = 0.5 vs Ha: p ≠ 0.5 If we do not reject Ho, we have found:

.6

Hypotheses for a test are given, as well as some information about the sample. Indicate where the randomization distribution will be centered. Hypotheses: Ho: p = 0.6 vs Ha: p > 0.6 Sample: p̂ = 0.77

Right-tail test

Hypotheses for a test are given, as well as some information about the sample. Indicate whether the test is a left-tail test, a right-tail test, or a two-tail test. Hypotheses: Ho: p = 0.8 vs Ha: p > 0.8 Sample: p̂ = 0.96

No

If the p-value is 0.060, are the results significant at a 5% level?

Do not reject Ho

If the p-value is 0.22, what is the conclusion of the test if we use a 5 % significance level?

0.001.

Null and alternative hypotheses for a test for a population proportion are given, as well as sample results. Use StatKey to calculate the p-value.Round your answer to three decimal places, if required. Hypotheses: Ho: p = 0.6 vs Ha: p ≠ 0.6 Sample: p̂ = 142/200 = 0.71 with n = 200

Reject Ho

State the conclusion of the test based on a p-value of 0.0001, if we use a 5 % significance level.

Do not reject Ho

State the conclusion of the test based on a p-value of 0.2881, if we use a 5 % significance level.

0.01

The figure below shows a randomization distribution based on 1000 simulated samples for testing Ho: μ equals 50 vs Ho: μ > 50. In each case, use the distribution to decide which value is closer to the p-value for the observed sample mean. (See: https://imgur.com/MYpCY6I) (a) The p-value for x̅ equals 69 is closest to: 0.01 or 0.24?

0.29

The figure below shows a randomization distribution based on 1000 simulated samples for testing Ho: μ equals 50 vs Ho: μ > 50. In each case, use the distribution to decide which value is closer to the p-value for the observed sample mean. (See: https://imgur.com/MYpCY6I) (b) The p-value for x̅ equals 53 is closest to: 0.12 or 0.29?

0.05

The figure below shows a randomization distribution based on 1000 simulated samples for testing Ho: μ equals 50 vs Ho: μ > 50. In each case, use the distribution to decide which value is closer to the p-value for the observed sample mean. (See: https://imgur.com/MYpCY6I) (c) The p-value for x̅ equals 62 is closest to: 0.05 or 0.45?

.001

The null and alternative hypotheses for a population proportion, as well as the sample results, are given. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Use at least 1000 samples. Hypotheses: Ho: p = 0.5 vs Ha: p < 0.5 Sample: p̂ = 35/100 = 0.35 with n = 100

0.102

The null and alternative hypotheses for a population proportion, as well as the sample results, are given. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Use at least 1000 samples. Hypotheses: Ho: p = 0.5 vs Ha: p > 0.5 Sample: p̂ = 30/50 = 0.6 with n = 50

p-value = 0.02

Two possible p-values are given for a test. Which one provides the strongest evidence against Ho p-value equals 0.02 or p-value equals 0.92