Statisitc Homework 4

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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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Answer the following : a) Simulate drawing one sample of size 50. Let’s treat this as a random sample of 50 people voters. What sample proportion ¯p did you get? Did you expect to see the sample proportion to be exactly the same as the population proportion p = 53.8%? Why / why not? I expect the proportion to be very similar but it wasn't quite exactly the same due to a random variation b) Keep the sample size n as 50 and population proportion p as 0.538, but now simulate drawing 40,000 samples of that size. (Click ”10,000” circled in the picture above then click ”Draw samples” four times). i) Submit the picture of the histogram of 40,000 sample proportions you generated in your HW answer. ii) Describe the shape, center and spread of this sampling distribution of sample proportion. (Hint: The app shows mean, standard deviation of sampling distribution).

ii. The shape is bell shaped which tends to be very normal for the distribution. Standard deviation is 0.0702 c) Use a formula we learned from Ch 7 to verify the mean and standard deviation of sampling distribution from part b). (Note: It is okay to see a small difference (+/-0.0002) between the calculated values using formulas and values from simulation.) p√( 1 - p ) / n = √0.54(1-0.54)/ 50 = 0.0705 d) Now change the sample size (n) to 200, keeping the population proportion 0.538. Simulation the exit poll at least 10,000 times. How did the sampling distribution of sample proportion change from part b)? Compare shape, mean, and spread when n = 200 vs n = 50.

From the change within the graph, it seemed like the bell-shaped distribution seemed to stay the same. The standard deviation did change to 0.0354 while the n= 50 stayed the same. Problem 2 Use the sampling distribution web app from Problem 1. This time use the population proportion p = 0.97, and n = 100. Simulate at least 10,000 samples. What is the shape of the sampling distribution of sample proportion? Is it approximately normal? If not, why not? (Hint: Check lecture notes page 99, Central Limit Theorem for ˆp) This is not a normal distribution because the is skewed a lot more to the left which makes this information biased than the others. CLT requires more information such as votes to be greater than 15 to say no. CLT: np > 15 yes n(1-p) > 15 no Problem 3 Rafe was diagnosed with high blood pressure. He was able to keep his blood pressure in control for several month by taking blood pressure medicine. Rafe’s blood pressure is monitored by taking three readings a day, in early morning, at midday, and in the evening. a) During this period, the probability distribution of his systolic blood pressure reading had a mean of 130 (µ = 130) and a standard deviation of 6 (σ = 6). If the successive observations behave like a random sample from this distribution, find the mean and standard deviation of the sampling distribution of the sample mean for the three observations each day. σ / √µ = 6/ √36 = 3.46 n = (130, 3.46) b) Suppose that the probability distribution (population distribution) of his blood pressure reading is normal. What is the shape of the sampling distribution? Why?

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Since the distribution is normal, this would most likely provide a bell shaped distribution. c) Refer to part b. Find the probability that the sample mean exceeds 140, which is considered problematically high. 140-130/3.46 = 2.89 1- P(z < 2.89) = 1- 0.9981 = 0.0019 = 0.19% d) Find the probability that Rafe’s systolic blood pressure reading from one randomly selected morning is greater than 140. P( x > 140 ) = (2.89 - 1) / 2 = 2.39 Problem 4 Refer to Problem 3 where the probability distribution of Rafe’s systolic blood pressure reading had a mean of 130 (µ = 130) and a standard deviation of 6 (σ = 6). This time we are interested in sample mean from 21 observations. a) If we repeat Problem 3 part c) to find P(X >¯ 140) but with sample mean from 21 observations (n = 21) instead of 3 observations, how does your answer change? Increase, decrease or remain the same? Explain. Mean = 6/√ 21 = 1.31 σ = 140 - 130/ 1.31 = 7.63 Since this probability distribution is greater than and/or equal to 140, it is 0% which indicated that this decreases because if the sampling size was the be bigger, than there are less opportunity or chances for it to be greater and succeed 140 . b) Use a 68 - 95- 99.7 rule to the sampling distribution of sample mean with (n=21) to find the interval that contains the middle 95% of the distribution. If we can’t apply the 68-95-99.7 rule, explain why.

q(norm) (0.975, 130, 60) = 141.7598 - q(norm(0.025, 130, 6) = 118.2402 Problem 5 Based on data from the 2010 Major League Baseball season, X = number of home runs the San Francisco Giants hit in a game has the population mean of 1.0 and population standard deviation of 1.0. a) Do you think X has a normal distribution? Why or why not? Yes, i think it is a normal distribution because it is symmetrical and x follows normal distribution. X ∼ N(µ, σ) σ = 1.0 µ = 1.0 b) Suppose that this year X has the same distribution. Report the shape, mean, and standard deviation of the sampling distribution of the mean number of home runs the team will hit in its 36 games. The normal distribution is symmetrical, bell-shaped distribution in which the mean, median and mode are all equal. Therefore, the shape of the distribution is symmetric about mean. Sample size = n = 36 Mean = µ = 1.0 Standard deviation= σ =1.0 Standard deviation - (x-1.0)/(1.0/ √36) = 1.0/6 = 0.1667 c) Based on the answer to part b), find the probability that the sample mean number of home runs per game from 36 games will exceed 1.25. Sample size = n = 36

Mean = 1.0 Standard deviation =1.0 =P(Z > 0.25/0.1667) = P(Z > 1.50) = 0.0668 The sample mean from 36 games would exceed 1.25 making equal 0.0668 Problem 6: Which of the following is not correct? Briefly explain why it is incorrect. The standard deviation of a statistic describes : (A) The standard deviation of the sampling distribution of that statistic **(B) The standard deviation of the individual observations in a sample data estimates - This is incorrect because it is all inclusive observations. (C) The variability in the values of the statistic for repeated random samples of the same size Problem 7: Which of the following is correct? For each incorrect option, briefly explain why it is incorrect. The sampling distribution of a sample mean for a random sample size of 100 describes : (A) How sample means tend to vary from a random sample to sample of size 100. - This is incorrect because this describes standard deviation. (B) How observations tend to vary from person to person in a random sample of size 100 - It does matter what varies upon from person to person. **(C) How the data distribution looks like the population distribution when the sample size is larger than 30. (D) How sample standard deviation (s) varies among samples of size 100. - No because standard deviation is about how variables change amongst the mean.

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