Math 227 Chapter 10 and 11 Review: Confidence Intervals and Hypothesis TestingKEY
1. What is a type I error in hypothesis testing? What is a type II error?
Type 1 error: Rejecting null when it should not have been rejected.
Type 2 error: Not rejecting the null when it should have been rejected.
2. A hypothesis test is conducted using two different significance levels. Computer technology gives us the following outputs.
z = 1.58 and z = 1.87
a)Which of the two results will give a smaller p-value for a right tail test? Draw a normal curve with the appropriate area shaded to support your answer.1.87
b) Which of the two cases would be considered more unusual (less likely to happen by chance)?1.87 is farther to the right resulting in a smaller area. The probability will be less and therefore more unusual.
3. When conducting a hypothesis test computer technology give a p-value of 0.063.
- At a significance level of , is the result unusual? No, 0.063 not less than 0.01
- At a significance level of , is the result unusual? No, 0.063 not less than 0.05
- At a significance level of , is the result unusual? Yes 0.063 is less than 0.10
4. A 2003 study of dreaming found that out of a random sample of 113 people, 92 reported dreaming in color. However, the rate reported in the 1940’s was 0.29.
a. = ? p = ?
b. Identify each of the symbols in part (a) as a statistic or a parameter.is a parameter (from population) and is a statistic (from a sample)
c. Check to see whether the conditions for using a one proportion z-test are met assuming the rate in the 1940’s is still true. Do not conduct a hypothesis test.
npq = (113)(0.29)(0.71) = 23.2667 23.2667 > 10 yes, condition is met.
Note: There is a huge difference in the two proportions. If a hypothesis test were to be conducted, it is likely that we would have evidence to support that the proportion of people who dream in color is greater that of the past which was 0.29. What could have caused this?
Well today we are exposed to more color images compared to the past: television, photographs, movies played at theatres, etc.
5. The mean age of all 2550 students at a small college is 22.8 years with a standard deviation of 3.2 years and the distribution is right skewed. A random sample of 4 students’ ages is obtained at the mean is 23.2 years with a standard deviation of 2.4 years.
a. σ = ? = ? s = ? µ = ? n = ?σ = 3.2 (standard deviation of population), = 23.2 (mean of sample), s = 2.4 (standard deviation of sample), µ = 22.8 (mean of population), n= 4 (sample size)
b. Identify each of the symbols in part (a) as a statistic or a parameter.σ = 3.2 (parameter, from population), = 23.2 (statistic, from sample), s = 2.4 (statistic, from sample), µ = 22.8 (parameter, from population)
c. Are the conditions satisfied for this example? No, the sample of 4 is less than 30 and it came from a population that was not normally distributed.
6. A teacher giving a true/false test wants to make sure her students do better than they would if they were simply guessing, so she forms a hypothesis to test this. Her null hypothesis is that a student will get 50% of the questions on the exam correct. The alternative hypothesis is that the student is not guessing and should get more than 50% in the long run.
Set up:
Ho: p= 0.50
Ha: p > 0.50
A student gets 30 out of 50 questions, or 60% correct. Computer technology gives a p-value of 0.079.
a) Explain the meaning of the p-value in context.There is a 7.9% chance that if the student just guessed, the student would get 30 out of 50 correct.
b) Would you reject the null hypothesis at a significance level of 5%?0.079 is not less than 0.05. Do not reject the null. In this case, not unusual to get 60% correct if the student was just guessing. (The conclusion would be that the student was probably guessing.)
c) Would you reject the null hypothesis at a significance level of 10%?0.079 < 0.10. Reject the null. In this case, it would be unusual to get 60% correct if the student was just guessing. (The conclusion would be that the student was probably not guessing.)
7. Judging on the basis of experience, a politician claims that 48% in Pennsylvania have voted for an independent candidate in past elections. Suppose you surveyed 20 randomly selected people in Pennsylvania, and 11 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in Pennsylvania that have voted for an independent candidate is 48%.
a) Calculate the test statistic.
Note: (The percent voting for an independent candidate is 48%)
Also, conditions were not satisfied since npq = 20(0.48)(0.52) = 4.992 which is not greater than 10.
But let’s continue for practice:
b) Based only on this test statistic, would you reject the null hypothesis that the overall proportion of voters is 48% based on a significance level of 5%?95% corresponds to a z value of roughly 2. Therefore,the z statistic was not unusual. It was not more than 2 SD’s from the population mean of 0.48.
8. A 2003 study of dreaming found that out of a random sample of 113 people, 92 reported dreaming in color. However, the rate reported in the 1940’s was 0.29. A researcher wanted to see if the proportion of dreaming in color has increased since the 1940’s. Let’s assume the necessary conditions are met. Perform a hypothesis test to test this claim. Use a 1% significance level.
A. Does this require a one-sample, two-sample, or paired t-test?One sample (proportion) since only one sample was taken.
B. Conduct a hypothesis test
a. State the hypothesis.
(The percentof people who dream in color is 29%.)
percentof people who dream in color is greater than 29%.)
b. Several outputs from computer technology are shown below. Circle the correct output for this hypothesis.
Write the p value:Result B: p value: less than 0.0001
Result A:
Result B:
Result C:
c. Do you reject or not reject the null hypothesis?P value is unusual (0.0001) and it is less than the significance level of 1% (0.01). Reject the null.
d. Write your conclusion.There is significant evidence to support the claim that the percent of people who dream in color is greater than 29%.
e. Computer technology has provided the following confidence interval. Interpret the confidence interval. Then explain how the confidence interval supports the conclusion from the hypothesis test.
CI: (0.742, 0.886) One can be 95% confident that the true population proportion of people who dream in color is between 74.2% and 88.6%.
Since the CI does not contain the population proportion of 29% than there is evidence that the proportion of people who dream in color is no longer 29%. In fact, the CI shows that it is greater.
9. The population mean height for a 3 year old boy in the U.S. is 38 inches. Suppose a random sample of 15 non-U.S. boys are measured and the mean height for this group is 37.2 inches with a standard deviation of 3 inches. The non-U.S. boys were independently sampled. Assume that heights are normally distributed in the population.
A. Have the conditions been met to perform a hypothesis test? Explain.Yes. Although the sample size is not greater than 30, it does come from a normally distributed population.
B. Does this require a one-sample, two-sample, or paired t-test?One sample (mean) since only one sample was taken.
C. Perform a hypothesis test to determine if the mean height for non-U.S. boys is less than the U.S. population mean height. Use a significance level of 0.05.
a. State the hypothesis
(The mean height for non U.S. 3 year old boys is 38 inches.)
(The mean height for non-U.S. 3 year old boys is less than 38 inches.)
b. Computer technology gives the following output.
c. Do you reject or not reject the null hypothesis?P value = 0.1596 which is not unusual and 0.1596 is not less than 0.05. Do not reject the null.
d. Write your conclusion.There is not enough evidence to support the claim that non U.S. 3 year old boys have a mean height less than 38 inches.
e. The 95% confidence interval is shown below.Interpret the confidence interval. Then explain how the confidence interval supports the conclusion from the hypothesis test.
95 % CI: (35.5 38.9). One is 95% confident that the true mean height of non U.S. 3 year old boys is between 35.5 and 38.9 inches.
Since the population mean height of 38 inches is in the CI, then there is no evidence that the mean has since changed or is now different. Therefore, there is no evidence to support the claim that non U.S. 3 year old boys have a mean height less than 38 inches.
This supports the conclusion from the hypothesis test.
10. A recent article (New England Journal of Medicine, 2010) reported the results of an experiment on reducing the likelihood that men develop prostate cancer. The investigators randomly assigned 3305 men to receive the drug Dutasteride and assigned 3424 men to receive a placebo. Of those receiving the drug, 659 developed prostate cancer. Of those men who received the placebo, 858 developed prostate cancer. A. Does this require a one-sample, two-sample, or paired t-test?two sample (proportion)
B. Find the percentage of men that developed prostate cancer in each group. Let p1 be the proportion who received the drug and p2 the proportion who received the placebo.
= 0.199 = 0.251
C. Perform a hypothesis test to determine if the drug made a difference inlowering prostate cancer using a 5% significance level. Assume the conditions have been met.
a. State the hypothesis.
(The proportion of cancer in the males who took the drug is the same as those who took the placebo.)
The proportion of cancer in the males who took the drug is less than those who took the placebo.)
OR
(There is no differencein the proportion between the males who took the drug and those who took the placebo.)
There is a differencein the proportion between the males who took the drug and those who took the placebo. (Males taking drug having a lower proportion of cancer.))
b. Several outputs from computer technology are shown below. Circle the correct output for this hypothesis.
Write the p value:Result B: p value is less than 0.0001.
Result A
Result B
Result C
c. Do you reject or not reject the null hypothesis?P value of 0.0001 is unusual and 0.0001 is less than the significance level of 5%. Reject the null.
d. Write your conclusion.There is sufficient evidence to support the claim that the cancer rate in the group taking the drugis less than the cancer rate in the group taking the placebo.
e. Computer technology has provided the following confidence interval. Interpret the confidence interval. Then explain how the confidence interval supports the conclusion from the hypothesis test.
The 95% CI is the correct CI. CI: (-0.071, -0.031) One is 95% confident that the true difference in the proportion of cancer rates is between – 7.1 % and – 3.1%. (The negative sign indicates that the cancer rate for the second group was higher. In this case, the placebo group.)
Since the difference of zero percent is not in the CI, then we can conclude that there was a difference in cancer rates. In fact, the CI shows that the cancer rate in the drug group was lower compared to the placebo group. This is the same conclusion from the hypothesis test.
11. A researcher wants to know if a daughter’s height is about the same as the mother’s height. Twenty randomly selected mothers and daughters were chosen to participate. The results are shown below.
A. Does this require a one-sample, two-sample, or paired t-test?Paired t test (two samples but paired: A daughter was paired with a mother.)
B. Let the mother’s height be the first sample and the daughter’s height be the second sample. Computer technology determines that the mean height for the mother is 63.15 inches with a standard deviation of 2.22 inches. The daughter’s mean height is 63.55 inches with a standard deviation of 2.24 inches.
C. Perform a hypothesis test to determine a daughter’s height is about the same as her mother’s height using a 5% significance level. Assume the conditions have been met.
a. State the hypothesis.
(The mean height between a mother and daughter is the same.)
(The mean height between a mother and daughter is not the same.)
OR
(There is no difference in mean height between a mother and daughter.)
(There is a difference in mean height between a mother and daughter.)
OR
(There is no difference in mean height between a mother and daughter.)
(There is a difference in mean height between a mother and daughter.)
b. The output from computer technology is given below.
Write the p value:0.1453
c. Do you reject or not reject the null hypothesis?0.1453 is not unusual and 0.1453 is not less than 5%. Do not reject the null.
d. Write your conclusion.There is not enough evidence to support the claim that there is a difference in mean height between a mother and daughter.
e. Computer technology has provided the following confidence interval. Interpret the confidence interval. Then explain how the confidence interval supports the conclusion from the hypothesis test.
95% CI: (- 0.951 0.151). One can be 95% confident that the true difference in mean height between a mother and daughter is between -0.951 and 0.151 inches.
Since the mean difference of zero inches is in the CI, then there is no evidence that there is a difference between the mean heights. This is the same conclusion from the hypothesis test.
12. A veterinarian wants to compare the weight of two breeds of dogs. He believes that German Shepherds are larger than Doberman Pinchers. A random sample of 20 male German Shepherds found that their average weight was 112 pounds with a standard deviation of 28 pounds. A random sample of 14 male Dobermans found that their average weight was 107 pounds with a standard deviation of 20 pounds. Assume the weights of are normally distributed and that the conditions have been satisfied.
I. Does this require a one-sample, two-sample, or paired t-test?Two sample (means, not paired)
II. Perform a hypothesis test to investigate the veterinarian’s claim that German Shepherds are larger than Doberman Pinchers. Use a 0.05 significance level. Assume the conditions have been met.
a. State the hypothesis (let weight of the German Shepherds be the first sample and the weight of the Dobermans be the second sample)
(The mean weight for German Shepherds is the same as the mean weight for Doberman Pinchers.)
(The mean weight for German Shepherds is greater than the mean weight for Doberman Pinchers.)
OR
(There is no difference in the mean weight between German Shepherds and Doberman Pinchers.)
(There is a difference in the mean weight between German Shepherds and Doberman Pinchers. (German Shepherds having a higher mean weight.))
OR
(There is no difference in the mean weight between German Shepherds and Doberman Pinchers.)
(There is a difference in the mean weight between German Shepherds and Doberman Pinchers. (German Shepherds having a higher mean weight.))
b. Computer technology gives the following output
Do you reject or not reject the null hypothesis?P value = 0.274 which is not unusual and 0.274 is not less than 0.05. Do not reject the null.
c. Write your conclusion.There is not enough evidence to support the claim that German Shepherds have a mean weight that is greater than Doberman Pincers.
d. Computer technology has provided the following confidence interval. Interpret the confidence interval. Then explain how the confidence interval supports the conclusion from the hypothesis test.
95% CI: (- 11.77, 21.77). One can be 95% confident that the true difference in mean weights is between -11.77 and 21.77 pounds.
Since the mean difference of zero pounds is in the CI then there is no difference in the mean weights.
This is the same conclusion from the hypothesis test.