1. A random sample of21wolf litters in Ontario, Canada, gave an average ofx₁=4.7wolf pups per litter, with estimated sample standard deviations₁=1.0.Another random sample of7wolflitters in Finland gave an average ofx₂=2.8wolfpups per litter, with sample standard deviations₂=1.4.(a) Find an 85% confidence interval for₁-₂,the difference in population mean litter size between Ontario and Finland. (Round your answers to one decimal place.)

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(b) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 85% level of confidence, does it appear that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland?Because the interval contains only positive numbers, we can say that the average litter size of wolf pups is greater in Ontario.Because the interval contains both positive and negative numbers, we can not say that the average litter size of wolf pups is greater in Ontario. We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers, we can say that the average litter size of wolf pups is greater in Finland. (c) Test the claim that the average litter size of wolf pups in Ontario is greater than the average litter size of wolf pups in Finland. Use= 0.01.(i) What is the level of significance? State the null and alternate hypotheses. H₀:₁₂;H₁:₁=₂H₀:₁=₂;H₁:₁₂ H₀:₁=₂;H₁:₁<₂H₀:₁=₂;H₁:₁>₂ (ii) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The Student'st. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations.The Student'st. We assume that both population distributions are approximately normal with unknown standard deviations. What is the value of the sample test statistic? Compute the correspondingzortvalue as appropriate. (Test the difference₁₂.Round your answer to three decimal places.) (iii) Find (or estimate) theP-value. P-value > 0.2500.125 <P-value < 0.250 0.050 <P-value < 0.1250.025 <P-value < 0.0500.010 <P-value < 0.025P-value < 0.010 Sketch the sampling distribution and show the area corresponding to theP-value.

(iv) Based on your answers to parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level? At the= 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the= 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the= 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the= 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence to suggest that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland.Reject the null hypothesis, there is insufficient evidence to suggest that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland. Reject the null hypothesis, there is sufficient evidence to suggest that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland.Fail to reject the null hypothesis, there is insufficient evidence to suggest that the average litter size of wolf pups in Ontario is greater than the average litter size in Finland.

2. A study considered the question, "Are you a registered voter?" Accuracy of response was confirmed by a check of city voting records. Two methods of survey were used: a face-to-face interview and a telephone interview. A random sample of92people were asked the voter registration question face to face. Of those sampled,seventy-ninerespondents gave accurate answers (as verified by city records). Another random sample of82people were asked the same question during a telephone interview. Of those sampled,seventy-fiverespondents gave accurate answers. Assume the samples are representative of the general population.(a) Letp₁be the population proportion of all people who answer the voter registration question accurately during a face-to-face interview. Letp₂be the population proportion of all people who answer the question accurately during a telephone interview. Find a 95% confidence interval forp₁-p₂. (Round your answers to three decimal places.)

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(b) Does the interval contain numbers that are all positive? all negative? mixed? Comment on the meaning of the confidence interval in the context of this problem. At the 95% level, do you detect any difference in the proportion of accurate responses from face-to-face interviews compared with the proportion of accurate responses from telephone interviews?Because the interval contains only positive numbers, we can say that there is a higher proportion of accurate responses in face-to-face interviews.Because the interval contains both positive and negative numbers, we can not say that there is a higher proportion of accurate responses in face-to-face interviews. We can not make any conclusions using this confidence interval.Because the interval contains only negative numbers, we can say that there is a higher proportion of accurate responses in telephone interviews. (c) Test the claim that there is a difference in the proportion of accurate responses from face-to-face interviews compared with telephone interviews. Use= 0.05.(i) What is the level of significance? State the null and alternate hypotheses. H₀:p₁=p₂;H₁:p₁<p₂H₀:p₁>p₂;H₁:p₁=p₂ H₀:p₁=p₂;H₁:p₁p₂H₀:p₁=p₂;H₁:p₁>p₂ (ii) What sampling distribution will you use? What assumptions are you making? The Student'st. We assume that both population distributions are approximately normal with known standard deviations.The Student'st. We assume that both population distributions are approximately normal with unknown standard deviations. The standard normal. We assume that both population distributions are approximately normal with known standard deviations.The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. What is the value of the sample test statistic? Compute the correspondingzortvalue as appropriate. (Test the differencep₁p₂.Round your answer to two decimal places.) (iii) Find (or estimate) theP-value. (Round your answer to four decimal places.) Sketch the sampling distribution and show the area corresponding to theP-value.

(iv) Based on your answers to parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level? At the= 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the= 0.05 level, we reject the null hypothesis and conclude the data are statistically significant. At the= 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the= 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (v) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is insufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews.Reject the null hypothesis, there is sufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews. Reject the null hypothesis, there is insufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews.Fail to reject the null hypothesis, there is sufficient evidence to suggest that the proportion of accurate responses from face-to-face interviews differs from the proportion for telephone interviews.