CH 10 HW Unfinished
If a 90% confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between p1 and p2 at the 90% confidence level?
There is no difference between p1 and p2.
For one binomial experiment, n1 = 75 binomial trials produced r1 = 30 successes. For a second independent binomial experiment, n2 = 100 binomial trials produced r2 = 50 successes. At the 5% level of significance, test the claim that the probabilities of success for the two binomial experiments differ. (a) Compute the pooled probability of success for the two experiments. (Round your answer to three decimal places.) (b) Check Requirements: What distribution does the sample test statistic follow? Explain. (c) State the hypotheses. (d) Compute p̂1 − p̂2. Compute the corresponding sample distribution value. (Test the difference p1 − p2. Do not use rounded values. Round your final answer to two decimal places.) (e) Find the P-value of the sample test statistic. (Round your answer to four decimal places.) (f) Conclude the test. (g) Interpret the results.
(a) 0.01 H0: p1 = p2; H1: p1 < p2 (b) The standard normal. The number of trials is sufficiently large. -1.86 (c) 0.0314 left-tailed graph (d) At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Fail to reject the null hypothesis, there is insufficient evidence that the proportion of adults that attended college who believe in extraterrestrials is higher than that of adults who did not attend college.
In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer. At five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below. Weather Station | 1, 2, 3, 4, 5 January | 137, 122, 128, 64, 78 April | 104, 113, 100, 88, 61 Does this information indicate that the peak wind gusts are higher in January than in April? Use 𝛼 = 0.01. (Let d = January − April.) (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? (Round your answer to three decimal places.) (c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (e) State your conclusion in the context of the application.
(a) 0.01 H0: 𝜇d = 0; H1: 𝜇d > 0; right-tailed (b) The Student's t. We assume that d has an approximately normal distribution. 1.252 d = (x1 - x2) d̄ = Σd / n sd = √Σd^2 - (Σd)^2/n / n - 1 t = d̄ − 𝜇d / sd/√n (c) 0.125 < P-value < 0.250 positive right-tailed graph (d) At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Fail to reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.
In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer. Is fishing better from a boat or from the shore? Pyramid Lake is located on the Paiute Indian Reservation in Nevada. Presidents, movie stars, and people who just want to catch fish go to Pyramid Lake for really large cutthroat trout. Let row B represent hours per fish caught fishing from the shore, and let row A represent hours per fish caught using a boat. The following data are paired by month from October through April. Use a 1% level of significance to test if there is a difference in the population mean hours per fish caught using a boat compared with fishing from the shore. (Let d = B − A.) (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? (b) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? (Round your answer to three decimal places.) (c) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (e) State your conclusion in the context of the application.
(a) 0.01 H0: 𝜇d = 0; H1: 𝜇d ≠ 0; two-tailed (b) The Student's t. We assume that d has an approximately normal distribution. 2.001 d = (x1 - x2) d̄ = Σd / n sd = √Σd^2 - (Σd)^2/n / n - 1 t = d̄ − 𝜇d / sd/√n (c) 0.050 < P-value < 0.100 two-tailed graph outside shaded (d) At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Fail to reject the null hypothesis, there is insufficient evidence to claim a difference in population mean hours per fish between boat fishing and shore fishing.
Based on information from a previous study, r1 = 33 people out of a random sample of n1 = 103 adult Americans who did not attend college believe in extraterrestrials. However, out of a random sample of n2 = 103 adult Americans who did attend college, r2 = 46 claim that they believe in extraterrestrials. Does this indicate that the proportion of people who attended college and who believe in extraterrestrials is higher than the proportion who did not attend college? Use 𝛼 = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. (b) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? (Test the difference p1 − p2. Do not use rounded values. Round your final answer to two decimal places.) (c) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (e) Interpret your conclusion in the context of the application.
(a) 0.05 H0: p1 = p2; H1: p1 < p2 (b) The standard normal. The number of trials is sufficiently large. -2.92 (c) 0.0017 left-tailed graph (d) At the 𝛼 = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. (e) Reject the null hypothesis, there is sufficient evidence that the proportion of trusting people in Chicago is higher in the older group.
A random sample of 49 measurements from one population had a sample mean of 13, with sample standard deviation 5. An independent random sample of 64 measurements from a second population had a sample mean of 16, with sample standard deviation 6. Test the claim that the population means are different. Use level of significance 0.01. (a) What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute x1 − x2 and the corresponding sample distribution value. (d) Estimate the P-value of the sample test statistic. (e) Conclude the test. (f) Interpret the results. (g) Find a 95% confidence interval for 𝜇1 − 𝜇2. Explain the meaning of the confidence interval in the context of the problem.
(a) Here we are not given the population standard deviations, so they are treated as unknown values. Therefore, we will use a t distribution for the sample test statistic. We have a random sample of 49 measurements from the first population, and an independent random sample of 64 measurements from the second population. We cannot assume that the populations from which these samples are drawn are approximately normal. The sample sizes are n1 = 49 ≥ 30 and n2 = 64 ≥ 30. We can therefore conclude that the requirements have been met to use a Student's t distribution for the sample test statistic. (b) H0: 𝜇1 = 𝜇2 H0: 𝜇1 ≠ 𝜇2 (c) We identify the values x̄1 = 13, x̄2 = 16, s1 = 5, and s2 = 6. The sample test statistic is x̄1 - x̄2 = -3. t = x̄1 - x̄2 /√s1^2/n1 + s2^2/n2 = -3 /√25/49 + 36/64 = -2.897 We will also need the degrees of freedom for this sample statistic, which is given by the smaller of n1 − 1 and n2 − 1. Since n1 − 1 = 48 and n2 − 1 = 63, we find that d.f. = 48. (d) Since the null and alternate hypotheses are H0: 𝜇1 − 𝜇2 = 0, H1: 𝜇1 − 𝜇2 ≠ 0, we are performing a two-tailed test Note that the t-value 2.897 is not an entry in the table. However, we can say that this value falls directly between table values 2.690 and 3.520 in the row for d.f. = 45. Due to the type of test we have, we will need to use the column headings for two-tail area to determine the P-value. Thus, the P-value is in the following interval. 0.0001 < P-value < 0.010 (e) We are testing at a level of significance of 0.01, so 𝛼 = 0.01. Comparing the interval 0.001 < P-value < 0.010 to the level of significance, we have that the P-value < 𝛼. Therefore, we reject the null hypothesis. (f) By rejecting H0, this means that the sample means are sufficiently different, so we can conclude at the 0.01 level of significance that the population means are different. (g) Using the table excerpt above, we can see that the critical value necessary for the calculation of this confidence interval is t0.95 = 2.014 E = tc√s1^2/n1 + s2^2/n2 = 2.014√5^2/49 + 6^2/64 = 2.086 = (x̄1 - x̄2) - E = (13 - 16) - 2.086 = -5.09 = (x̄1 - x̄2) + E = (13 - 16) + 2.086 = -0.91 The confidence interval contains all negative numbers. This indicates that at the 95% confidence level, 𝜇1 is less than 𝜇2.
A random sample of 49 measurements from a population with population standard deviation 𝜎1 = 3 had a sample mean of x̄1 = 8. An independent random sample of 64 measurements from a second population with population standard deviation 𝜎2 = 4 had a sample mean of x̄2 = 10. Test the claim that the population means are different. Use level of significance 0.01. (a) What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute x1 − x2. x̄1 − x̄2 = _____ Compute the corresponding sample distribution value. (Test the difference 𝜇1 − 𝜇2. Round your answer to two decimal places.) (d) Find the P-value of the sample test statistic. (Round your answer to four decimal places.) (e) Conclude the test. (f) Interpret the results. (g) Find a 99% confidence interval for 𝜇1 − 𝜇2 (Round your answers to two decimal places.) lower limit _____ upper limit ______ Explain the meaning of the confidence interval in the context of the problem.
(a) The standard normal distribution. Samples are independent, the population standard deviations are known, and the sample sizes are sufficiently large. (b) H0: 𝜇1 = 𝜇2; H1: 𝜇1 ≠ 𝜇2 (c) x̄1 - x̄2 = -2 -3.04 z = x̄1 - x̄2 / √𝜎1^2/n1 + 𝜎2^2/n2 (d) 0.0024 P-value = 2P(-3.04 < z < 3.04) (e) At the 𝛼 = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. (f) Reject the null hypothesis, there is sufficient evidence that there is a difference between the population means. (g) lower limit = -3.7 (x̄1 - x̄2) - E upper limit = -0.3 (x̄1 - x̄2) + E E = zc√𝜎1^2/n1 + 𝜎2^2/n2 At the 99% level of confidence, it appears that the difference between population means is between the lower limit and the upper limit.
For a random sample of 36 data pairs, the sample mean of the differences was 1.16. The sample standard deviation of the differences was 3. At the 5% level of significance, test the claim that the population mean of the differences is different from 0. (a) Is it appropriate to use a Student's t distribution for the sample test statistic? Explain. What degrees of freedom are used? (b) State the hypotheses. (c) Compute the t value. (d) Estimate the P-value of the sample test statistic. (e) Do we reject or fail to reject the null hypothesis? Explain. (f) What do your results tell you?
(a) We cannot assume that the data is approximately normal. The sample size n is greater than or equal to 30. We can therefore conclude that the requirements have been met to use a Student's t distribution for the sample test statistic d̄. d.f. = n - 1 = 36 - 1 = 35 (b) H0 : 𝜇d = 0 H1 : 𝜇d ≠ 0 (c) We have a random sample of 36 data pairs where the sample mean of the differences is 1.16 and the sample standard deviation of the differences is 3. Then the sample test statistic is d̄ = 1.16. Using these values, along with 𝜇d = 0, we calculate the corresponding t value for the test statistic. t = d̄ − 𝜇d / sd/√n = 1.16 - 0 / 3/√36 = 2.320 (d) The P-value for the sample statistic t = 2.320 is determined using a Student's t distribution table. Since the null and alternate hypotheses are H0: 𝜇d = 0, H1: 𝜇d ≠ 0, we are performing a two-tailed test. Note that the t value 2.320 is not an entry in the table. However, we can say that this value falls directly between table values 2.030 and 2.438 in the row for d.f. = 35. Due to the type of test we have, we will need to use the column headings for two-tailed area to determine the P-value. Thus, the P-value is in the following interval. 0.020 < P-value < 0.050 (e) We are testing at a 5% level of significance, so 𝛼 = 0.05. Comparing the interval 0.020 < P-value < 0.050 to the level of significance, we have that the P-value < 𝛼. Therefore, we reject reject the null hypothesis. (f) By rejecting H0, this means that the sample mean of differences is sufficiently different from 0, so we can conclude at the 5% level of significance that the population mean of differences is not 0.
Generally speaking, would you say that most people can be trusted? A random sample of n1 = 258 people in Chicago ages 18-25 showed that r1 = 42 said yes. Another random sample of n2 = 273 people in Chicago ages 35-45 showed that r2 = 73 said yes. Does this indicate that the population proportion of trusting people in Chicago is higher for the older group? Use 𝛼 = 0.05. (a) What is the level of significance? State the null and alternate hypotheses. (b) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? (Test the difference p1 − p2. Do not use rounded values. Round your final answer to two decimal places.) (c) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (e) Interpret your conclusion in the context of the application.
(a) answer answer (b) answer answer (c) answer answer (d) answer (e) answer
Alisha is conducting a paired differences test for a "before (B score) and after (A score)" situation. She is interested in testing whether the average of the "before" scores is higher than that of the "after" scores. (a) To use a right-tailed test, how should Alisha construct the differences between the "before" and "after" scores? (b) To use a left-tailed test, how should she construct the differences between the "before" and "after" scores?
(a) d = B - A (b) d = A - B
Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences that are described at length in the book Manual: A Guide to the Development and Use of the Myers-Briggs Type Indicator, by Myers and McCaulley.† Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. A random sample of 377 married couples found that 295 had two or more personality preferences in common. In another random sample of 568 married couples, it was found that only 24 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 99% confidence interval for p1 - p2. (Round your answers to three decimal places.) lower limit _____ upper limit _____ (b) Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the 99% confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?
(a) lower limit = 0.681 upper limit = 0/799 (b) Because the interval contains only positive numbers, we can say that a higher proportion of married couples have two or more personality preferences in common.
An article reported the results of a peer tutoring program to help children learn to read. In the experiment, Form 2 of the Gates-MacGintie Reading Test was administered to both an experimental group and a control group after 6 weeks of instruction, during which the experimental group received peer tutoring and the control group did not. For the experimental group n1 = 30 children, the mean score on the vocabulary portion of the test was x1 = 368.4, with sample standard deviation s1 = 39.3. The average score on the vocabulary portion of the test for the n2 = 30 subjects in the control group was x2 = 349.0 with sample standard deviation s2 = 56.2. (a) Use a 1% level of significance to test the claim that the experimental group performed better than the control group. (i) What is the level of significance? State the null and alternate hypotheses. (ii) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference 𝜇1 − 𝜇2. Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (v) Interpret your conclusion in the context of the application. (b) Find a 98% confidence interval for 𝜇1 − 𝜇2 (Round your answers to two decimal places.) lower limit _____ upper limit _____ Explain the meaning of the confidence interval in the context of the problem.
(ai) 0.01 H0: 𝜇1 = 𝜇2; H1: 𝜇1 > 𝜇2 (aii) The Student's t. Both sample sizes are large with unknown standard deviations. 1.549 (aiii) 0.050 < P-value < 0.125 right-tailed graph (aiv) At the 𝛼 = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (av) Fail to reject the null hypothesis, there is insufficient evidence that the mean score for the experimental group is higher than for the control group. (b) lower limit = -11.43 (x̄1 - x̄2) - E upper limit = 50.23 (x̄1 - x̄2) + E Because the interval contains both positive and negative numbers, this indicates that at the 98% confidence level, we cannot say that the population mean score for the experimental group was higher than for the control group.
REM (rapid eye movement) sleep is sleep during which most dreams occur. Each night a person has both REM and non-REM sleep. However, it is thought that children have more REM sleep than adults.† Assume that REM sleep time is normally distributed for both children and adults. A random sample of n1 = 9 children (9 years old) showed that they had an average REM sleep time of x1 = 2.9 hours per night. From previous studies, it is known that 𝜎1 = 0.5 hour. Another random sample of n2 = 9 adults showed that they had an average REM sleep time of x2 = 2.2 hours per night. Previous studies show that 𝜎2 = 0.7 hour. (a) Do these data indicate that, on average, children tend to have more REM sleep than adults? Use a 1% level of significance. (i) What is the level of significance? State the null and alternate hypotheses. (ii) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding z or t-value as appropriate. (Test the difference 𝜇1 − 𝜇2. Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (v) Interpret your conclusion in the context of the application. (b) Find a 98% confidence interval for 𝜇1 − 𝜇2 (Round your answers to two decimal places.) lower limit _____ upper limit _____ Explain the meaning of the confidence interval in the context of the problem.
(ai) 0.01 H0: 𝜇1 = 𝜇2; H1: 𝜇1 > 𝜇2 (aii) The standard normal. We assume that both population distributions are approximately normal with known standard deviations. 2.44 z = x̄1 - x̄2 / √𝜎1^2/n1 + 𝜎2^2/n2 (aiii) 0.0073 P-value = P(z > 2.44) right-tailed graph (aiv) At the 𝛼 = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. (av) Reject the null hypothesis, there is sufficient evidence that the mean REM sleep time for children is more than for adults. (b) lower limit = -3.7 (x̄1 - x̄2) - E upper limit = -0.3 (x̄1 - x̄2) + E E = zc√𝜎1^2/n1 + 𝜎2^2/n2 Because the interval contains only positive numbers, this indicates that at the 98% confidence level, the population mean REM sleep time for children is greater than that for adults.
Education influences attitude and lifestyle. Differences in education are a big factor in the "generation gap." Is the younger generation really better educated? Large surveys of people age 65 and older were taken in n1 = 36 U.S. cities. The sample mean for these cities showed that x1 = 15.2% of the older adults had attended college. Large surveys of young adults (age 25-34) were taken in n2 = 35 U.S. cities. The sample mean for these cities showed that x2 = 18.7% of the young adults had attended college. From previous studies, it is known that 𝜎1 = 6.2% and 𝜎2 = 4.8%. (a) Does this information indicate that the population mean percentage of young adults who attended college is higher? Use 𝛼 = 0.05. (i) What is the level of significance? State the null and alternate hypotheses. (ii) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference 𝜇1 − 𝜇2. Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (v) Interpret your conclusion in the context of the application. (b) Find a 90% confidence interval for 𝜇1 − 𝜇2 (Round your answers to two decimal places.) lower limit _____ upper limit _____ Explain the meaning of the confidence interval in the context of the problem.
(ai) 0.05 H0: 𝜇1 = 𝜇2; H1: 𝜇1 < 𝜇2 (aii) The standard normal. We assume that both population distributions are approximately normal with known standard deviations. -2.66 z = x̄1 - x̄2 / √𝜎1^2/n1 + 𝜎2^2/n2 (aiii) 0.0039 P-value = P(z > 2.66) left-tailed graph (aiv) At the 𝛼 = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. (av) Reject the null hypothesis, there is sufficient evidence that the mean percentage of young adults who attend college is higher. (b) lower limit = -5.66 (x̄1 - x̄2) - E upper limit = -1.34 (x̄1 - x̄2) + E E = zc√𝜎1^2/n1 + 𝜎2^2/n2 Because the interval contains only negative numbers, this indicates that at the 90% confidence level, the population mean percentage of young adults who attend college is higher than that of older adults.
An article reported the results of a peer tutoring program to help children learn to read. In the experiment, the children were randomly divided into two groups: the experimental group received peer tutoring along with regular instruction, and the control group received regular instruction with no peer tutoring. There were n1 = n2 = 30 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 49.3. For the control group, the mean score on the same test was x2 = 354.2, with sample standard deviation s2 = 50.5. (a) Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began. (i) What is the level of significance? State the null and alternate hypotheses. (ii) What sampling distribution will you use? What assumptions are you making? What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference 𝜇1 − 𝜇2. Round your answer to two decimal places.) (iii) Find (or estimate) the P-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level 𝛼? (v) Interpret your conclusion in the context of the application. (b) Find a 95% confidence interval for 𝜇1 − 𝜇2 (Round your answers to two decimal places.) lower limit _____ upper limit _____ Explain the meaning of the confidence interval in the context of the problem.
(ai) 0.05 H0: 𝜇1 = 𝜇2; H1: 𝜇1 ≠ 𝜇2 (aii) The Student's t. Both sample sizes are large with unknown standard deviations. -0.753 (aiii) 0.250 < P-value < 0.500 two-tailed graph (aiv) At the 𝛼 = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (av) Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean vocabulary scores between the control and experimental groups. (b) lower limit = -36.04 (x̄1 - x̄2) - E upper limit = 16.65 (x̄1 - x̄2) + E Because the interval contains both positive and negative numbers, this indicates that at the 95% confidence level, we cannot say that the population mean scores are different before any instruction began.
Consider a hypothesis test of difference of proportions for two independent populations. Suppose random samples produce r1 successes out of n1 trials for the first population and r2 successes out of n2 trials for the second population. What is the best pooled estimate p for the population probability of success using H0: p1 = p2?
(r1 + r2) / (n1 + n2)
Consider a hypothesis test of difference of means for two independent populations x1 and x2. What are two ways of expressing the null hypothesis?
H0: 𝜇1 = 𝜇2 or H0: 𝜇1 - 𝜇2 = 0
Consider two independent populations for which the proportion of successes in the first population is p1 and the proportion of successes in the second population is p2. What alternate hypothesis would indicate that the proportion of successes p1 in the first population is larger than the proportion of successes p2 in the second population? Express the alternate hypothesis in two ways.
H1: p1 > p2 or H1: p1 - p2 > 0
When conducting a test for the difference of means for two independent populations x1 and x2, what alternate hypothesis would indicate that the mean of the x2 population is larger than that of the x1 population? Express the alternate hypothesis in two ways.
H1: 𝜇1 < 𝜇2 or H1: 𝜇1 - 𝜇2 < 0
When testing the difference of means for paired data, what is the null hypothesis?
Ho: 𝜇d = 0
Consider a set of data pairs. What is the first step in processing the data for a paired differences test? What is the symbol for the sample test statistic? Describe the value of the sample test statistic.
Take the paired differences. The test statistic symbol is d̄, which is the mean of the paired differences.
When conducting a paired differences test, what is the value of n?
The number of data pairs.
Consider a hypothesis test of difference of means for two independent populations x1 and x2. Suppose that both sample sizes are greater than 30 and that you know 𝜎1 but not 𝜎2. Is it standard practice to use the normal distribution or a Student's t distribution?
Use the Student's t distribution because we do not know 𝜎2.
Are data that can be paired independent or dependent?
dependent
When using a Student's t distribution for a paired differences test with n data pairs, what value do you use for the degrees of freedom?
n - 1
Consider a hypothesis test of difference of proportions for two independent populations. Suppose random samples produce r1 successes out of n1 trials for the first population and r2 successes out of n2 trials for the second population. What is the sample test statistic for the test?
p̂1 - p̂2
Consider a hypothesis test of difference of means for two independent populations x̄1 and x̄2. What is the sample test statistic?
x̄1 - x̄2
