STATS Homework 7

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University of Minnesota-Twin Cities *

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Statistics

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Jan 9, 2024

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Problem 1 For each part below answer (i) state the name of relevant hypothesis test: • One sample z -test for proportion p - Normal hypothesis • One sample t -test for mean µ - t-test • Matched pairs t -test for mean of difference µ D - Two-sided hypothesis test • Independent two sample t -test for difference of means µ 1 − µ 2 - Two sided hypothesis test (ii) define parameter(s) of interest and state the null and alternative hypotheses. a) It is known that 40% of college students in the U.S. identified Instagram as their favorite social media application. To test if this information is still accurate in 2022, you surveyed 100 college students randomly and found 32% of them identified Instagram as their favorite. ➔ H0 : µ = 0.40 ➔ Ha : µ > 0.40 b) Forty-four sixth graders were randomly selected from a school district. Then, they were divided into 22 matched pairs, each pair having equal IQ’s. One member of each pair was randomly selected to receive special training. Then all of the students were given an IQ test. The researcher wants to test if the special training improves IQ scores. ➔ H0 : µ D = 0 ➔ Ha : µ D ≠ 0 c) An experiment investigated whether cell phone use increases drivers’ reaction times, using a sample of 100 college students. Students were randomly assigned to a cell phone group or to a control group, 50 to each. In a simulation of driving situations, a target flashed red or green at irregular periods. Participants pressed a brake button as soon as they detected a red light. The control group listened to the radio or audiobook while they performed the simulated driving. The cell phone group carried out a phone conversation about a political issue with someone in a separate room. The experiment measured each group’s mean response time over many trials. ➔ H0 : µ = .50 ➔ Ha : µ > .50 d) You want to know if students at the University of Minnesota sleeps more than 7 hours on average. To investigate this, you sampled a few class mates and asked how many hours they sleep on a typical week day. ➔ H0 : µ = .70 ➔ Ha : µ > .70 e) Lucy wanted to test if the selling prices of e-book is cheaper than prices of hard cover book. She first took a random sample of 7 books. Then she looked up e-book prices and hard copy prices of those 7 books. ➔ H0 : µ = 7 ➔ Ha : µ ≠ 7 Problem 2

98.6 F is commonly referred as the average body temperature. Is this information accurate? A study was conducted in 1992 to examine this belief. The oral body temperatures of a random sample of 100 healthy adults were measured. The mean and standard deviations are ¯ x = 98 . 2 F and s = 0 . 7 F . Use this sample information to test whether the true human body temperature is equal to 98.6 or not. Use α = 0 . 05. • Assumption : the sample size is large enough, greater than 100. Second, the observations are independent from each other. Lastly, the data is randomly sampled from the population of interest. This follows normal distribution. • Hypotheses : Ho: μ = 98.6 Ha: μ ≠ 98.6 • Test statistic: What is the distribution of test statistics? Based on the sample information provided in Problem 2, what is the value of the test statistic? (98.2 - 98.6) / 0.7√100 = -5.71 • P-value: Interpret the p-value in the context of the problem. ❖ 2 * pt (-5.71, df= 99, lower.tail = FALSE) • Conclusion and interpretation in context. ❖ Since the P value is less than the significance level, we must reject the null hypothesis, H0. This concludes that there is enough evidence to support that the human body temperature is not equal to 98.6. Problem 3: type 1 vs type 2 error A researcher wants to see if a new drug with potentially dangerous side effects is significantly better than the current drug. If it is found to be more effective, it will be prescribed to millions of people. a) What does a Type I error mean in this scenario? ❖ This showcases the probability of accepting the new drug would be better than the current drug. b) What does a type II error mean in this scenario? ❖ This is the possibility of rejecting that the new drug is going to be better than the current drug that is given. c) Which type of error is worse to make in this situation? Explain. ❖ Type 1 is the worst error because if the new drug is not better than the current drug, then that falls into many categories of it being false to millions of people. Problem 4 A graduate teaching assistant collected data to investigate whether study time per week (average number of hour) differed between students who planned to go to graduate school and those who did not. The data were as follows: Graduate school : 15, 7, 15, 10, 5, 5, 2, 3, 12, 16, 15, 37, 8, 14, 10, 18, 3, 25, 15, 5, 5 No graduate school : 6, 8, 15, 6, 5, 14, 10, 10, 12, 5

Use the following R command to input data. grad<-c(15, 7, 15, 10, 5, 5, 2, 3, 12, 16, 15, 37, 8, 14, 10, 18, 3, 25, 15, 5, 5) noGrad<-c(6, 8, 15, 6, 5, 14, 10, 10, 12, 5) a) Use R to find sample mean and sample standard deviation for each group. Copy and paste your R output. ❖ > mean(grad) ❖ [1] 11.66667 ❖ > sd(grad) ❖ [1] 8.338665 ❖ > mean(noGrad) ❖ [1] 9.1 ❖ > sd(noGrad) ❖ [1] 3.695342 b) Show formula with numbers plugged in to verify that standard error for difference between the sample means is 2.16. Interpret this value. √ 8.33^2/ 21 +3.70/ 10^2 = T = .005 c) What is the value of t-multiplier, t α/ 2 ,df =min( n 1 − 1 ,n 2 − 1) , to construct a 99% confidence interval? Copy and paste your R command and output. ❖ 9t(.005, df = 10 -1) = -3.25 d) Show formula with numbers plugged in to verify that 99% confidence interval to estimate the difference between the two population mean is (-4.4, 9.59). Interpret. mean(grad)-mean(noGrad)+-3.249836*2.161 = -4.4 mean(grad)-mean(noGrad)--3.249836*2.161 = 9.59 This means that we are 99% confident that the difference between hour average for those who plan to go to graduate school and those who will not is between these two points. (-4.4, 9.59) Problem 5 Use the same information from Problem 4 to conduct a hypothesis test to test whether the average study hours for students who plan to go to graduate school is greater than average study hours for students who don’t plan to go to graduate school. Use α = 0 . 1. (Do not use t.test()) • Assumption : assume whether the average study hours for students who plan to go to graduate school match those who do not plan to go. • Hypotheses : H0 : mu 1 = mu2 Ha: mu1 > mu2 • Test statistic : What is the distribution of test statistic? Based on the sample information provided in the Problem 4, what is the value of the test statistic? • P-value > pt(-3.25, df= .99) [1] 0.09602097 • Conclusion and interpretation in context. Since this has concluded that the p value is smaller than 0.1, we fail to reject the null hypothesis .

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Problem 6 Type two because it could have been due to not rejecting the null hypothesis due to it being false. Problem 7 Use R command t.test() to repeat Problem 5. (Test whether students who plan to go to graduate school study more than students who don’t at α = 0 . 1). Copy and paste your R command and output. Note to students : R uses ”Welch’s degrees of freedom” instead of minimum of n 1 − 1, n 2 − 1, hence gives a slightly different p- value from Problem 5. Problem 8 P-value of hypothesis test H a : µ 1 ̸ = µ 2 is 0.03. Which of the following is always true? Explain. (A) Using the same sample information, the 95% confidence interval for µ 1 − µ 2 will contain 0. (B) Using the same sample information, the 99% confidence interval for µ 1 − µ 2 will contain 0. (C) Using the same sample information, the 95% confidence interval for µ 1 −µ 2 will contain positive numbers only. P value is greater than 0.01 so we fail to reject h0, thus having the 99 percent interval contain 0.