HW5Problems

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Jan 9, 2024

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Problem 1 Part 1 In a clinical study, 1000 subjects were vaccinated with a vaccine manufactured with a new ingredient. Over a period of roughly 28 weeks, 6 of these subjects developed the flu. a) Find the point estimate of the population proportion that were vaccinated but still developed the flu. Specify its value and statistical notation for this estimate. P = 6/1000 = 0.006 of the subjects still developed the flu after getting a vaccination b) Show set up (formula with numbers plugged in) and verify that the standard error of this estimate is 0.0024. Briefly explain what this number means. Standard error = σp = √ p(1-p)/n = √ 0.006 (1-0.006)/1,000 = 0.0024 This is supposed to represent the spread in the distribution of the sampled population. c) What is the value of z-multiplier z α/ 2 for a 98% confidence interval? Find the margin of error for a 98% confidence interval. Round your margin of error to the nearest four decimal number. P (Z < 2.326) = .99 Moe = z * standard error = 2.326 * 0.0024 = 0.0056 Problem 1 Part 2 d) Use your answers from Problem 1 Part 1 and construct a 98% confidence interval. Interpret. Confidence interval = p - moe, p + moe (0.006 - 0.0056), (0.006 + 0.0056) (0.0004 , 0.0116) This is supposed to show that we are 98% confident that the population proportion of those who have developed the flu falls in this given interval. e) Based on your confidence interval, is it plausible that less than 1% of all people vaccinated with the vaccine will develop the flu? Explain. No, this is plausible because that values that are given here end up being greater than 1%. Therefore, it states that we cannot conclude that less than 1% would get the flu since the range follows from 4% to 1.16% Problem 2 Refer to the Problem 1.

Does the information provided in Problem 1 satisfy conditions (assumptions) of the large number confidence interval for p ? State two assumptions and determine whether each is met or not. (Hint: See lecture notes on page 106. There are two assumptions.) Yes , problem 1 had met the assumptions. Two assumptions are listed as randomized when there seems to be no known standard deviation or mean. This was met due to a random sample of 6 people out of 1000 that had gotten the flu after their shot. Normality is when the distribution seems to be normal and marks satisfaction in size, independence, and random sampling. This was also met due to the size of the sampling. Problem 3 An election is expected to be close. Pollsters planning an exit poll want their estimates be to within ± 0 . 02 with 90% confidence. How large should be the sample size if they have no prior information? Find Zα/2 for 90% confidence. 90% written as a decimal is 0.90. 1 – 0.90 = 0.10 = α and α/2 = 0.10/2 = 0.05 = p ME = Zα/2 * √ p(1-p)/n Sample size = n N = Z^2α/2 * √ p(1-p)/me^2 N = 1.64^2 * 0.5 (1 - 0.5)/0.02^2 = 1.64^2 * 625 = 1681 would be the sample size Problem 4 Part 1 The 2012 General Social Survey asked, “What do you think is the ideal number of children for a family to have?”. The 590 females who gave a numeric response from 0 to 6 had a median of 2, mean of 2.56, and standard deviation of 0.84. a) What is the point estimate of the population mean? Specify its value and statistical notation for this estimate. Point estimate = Sample mean = 2.56 b) Find the standard error of this estimate. Round your answer to the nearest four decimal number. Explain what this number means. Standard error = 0.84/ √590 = 0.0346 This number explains how it differs from the sample mean.

c) What is the value of t-multiplier t α/ 2 ,n− 1 for a 99% confidence interval? Find the margin of error for a 99% confidence interval. Round your margin of error to the nearest two decimal number. T a/2 (590-1) = t589 = 2.584 margin error = 0.0346*2.584=0.0894 ME = 0.09 Problem 4 Part 2 d) 99% confidence interval is (2.47, 2.65). Interpret the interval in context. This means that we are 99% confident that the mean number of children per female falls between 2.47 and 2.65 interval. e) Is it plausible that the population mean of females’ ideal number of children is greater than 2? No, ideally it is no plausible because the value outside of the confidence interval is greater than 2 Problem 5 Refer to Problem 4. State assumptions of confidence interval for population mean. Determine whether each assumption is met based on Problem 4 description. Explain. Since the mean and standard deviation had value, it was safe to assume that it was normal. The assumption of this interval is normal because the number or responders was greater than 30. Based of the number of respondents, it was met due to the volume given. Problem 6: Multiple Choice Question The following interprets the 99% confidence interval for population mean of females’ ideal number of children from Problem 4 part d) (2.47, 2.65). Which of the following is correct? (A) In the entire population, 99% of females’ ideal number of children fall within 2.47 and 2.65. (B) We are 99% confident the sample mean of females’ ideal number of children falls between 2.47 and 2.65. (C) If we repeatedly sample 590 females randomly and construct 99% confidence intervals, then in the long run, 99% of those intervals contain the population mean of females’ idea number of children. (D) There is 99% probability population mean is between 2.47 and 2.65.

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R Problem In this problem, we will use the class survey data set. We will assume that the population of interest is all students at the University of Minnesota and the variable of interest is average weekday sleep hours. Assume that the survey data represents a random sample of 391 students at the University of Minnesota. Use the following R code to import data from the survey data. survey_f22<-read.csv("http://users.stat.umn.edu/~parky/Fall2022Survey.csv", header=TRUE) a) Construct a histogram and Q-Q plot of students’ weekday sleep hours. Include both plots in your submission. Describe the shape of the distribution.

b) (Multiple Choice Question) The histogram from the previous part is : (i) a population distribution (ii) a data distribution (sample distribution) (iii) a sampling distribution Briefly explain. Data distribution observes distribution of the data that you have collected while sampling distribution focuses more on the wider aspect of data distribution that is randomly observed. c) State the assumptions of confidence interval for mean. Determine whether our sample satisfy this assumption This assumption is going to be Normal. Sample size needs to be very large and this sample size does satisfy the assumption because the sample size exceeded the number of 30 in order for it to be sufficient . d) Use t.test() command to construct a 90% confidence interval to estimate the population mean weekday sleep hours. Submit R command and output. Interpret the interval.

HW5Problems

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