Stat200_Fall Answer the following questions showing all work. Full credit will not be given to answers without work shown. If you use Minitab Express include the appropriate output (copy + paste) along with an explanation. Output without an explanation will not receive full credit. Round all answers to 3 decimal places. If you have any questions, post them to the Lesson 6 discussion board. 1. According to U.S. News and World Reports, 18.60% of all Penn State World Campus students have military experience. In a random sample of 60 students enrolled in World Campus sections of STAT 200, 10 reported having military experience. [30 points] A. Compute the sample proportion (p-hat) for this sample of n=60 STAT 200 students. The sample proportion (p-hat) = x/n = 10/60 = 0.1667 B. Should the exact method or normal approximation method be used to construct a sampling distribution in this situation? Explain your reasoning. Since, np and nq >= 10, normal approximation method should be used here B. Using StatKey, construct a sampling distribution for p=0.1860 and n=60. Generate at least 5,000 samples. Take a screenshot of your sampling distribution and paste it here. D. If the population proportion is 18.60%, what is the probability of taking a random sample of n=60 and finding a sample proportion more extreme than the one observed in this sample? Stat200_Fall Use StatKey to determine this proportion. Include a screenshot of your sampling distribution with this proportion highlighted. E. Given your results from part D, do you think that the proportion of all STAT 200 students who have military experience is different from the overall population of World Campus students where p=0.1860? Explain your reasoning. 2. For the following questions, assume a normally distributed population. [20 points] A. Given =50, =15, and n=25, compute the standard error of the mean. The standard error of the mean = /Sqrt(n) =15/Sqrt(25) = 15/5 = 3 B. Given =50, =15, and n=500, compute the standard error of the mean. The standard error of the mean = /Sqrt(n) =15/Sqrt(500) = 0.67082 C. Given =50, =3, and n=25, compute the standard error of the mean. The standard error of the mean =sigma/sqrt(n) =3/sqrt(25)= 0.6 D. Given =500, =15, and n=25, compute the standard error of the mean. The standard error of the mean = /Sqrt(n) = 15/sqrt(25)= 3 E. How does the standard error of the mean change when the population mean, population standard deviation, and sample size change? When Population mean increase or decrease then standard error will not change. It is equal to population standard deviation/root (sample size) so it increases with increase in population standard deviation and decreases with increase in sample size. 3. Using StatKey you are going to construct a sampling distribution for a mean. Select a population distribution that is NOT normal (e.g., the built in Hollywood Movies, Rock Bands, or Baseball Players datasets, or a skewed distribution of your own). Using the same population distribution for each, construct the distribution of sample means for N=4 and N=30. Take at least 5,000 samples. [20 points] A. Include a screen shots of your parent population and your two sampling distributions here. Stat200_Fall B. How are your two distributions of sample means similar? How are they different? C. Describe how your results relate to the Central Limit Theorem. Stat200_Fall 4. In the population ACT scores are normally distributed with a mean of 18 and a standard deviation of 6. Suppose that we are taking a simple random sample of 40 students from one high school. [30 points] A. Calculate the standard error of the mean. Standard Error = /sqrt(n) = 6/sqrt(40) = 0.9487 B. If we were to repeatedly pull samples of 40 individuals from the population of all ACT test takers, the distribution of sample means would have a mean of _18___ and a standard deviation of _0.9487___. C. Given the values from part B, 95% of random samples of n=40 will have sample means between ___ and ___. at = 0.05, Zcritical = 1.645 Margin of error = Zcritical x xx = 1.645 x 0.94868 = 1.56058 CI = [ + ME] = 18 + 1.56058 = [16.43942, 19.56058] D. What is the probability that you would pull a random sample of 40 individuals from the population of all test takers and they would have a sample mean of 19 or higher? z = (xx - )/xx = (19 - 18)/0.94868 = 1.054 P(z > 1.054) = 0.9488 E. Suppose that the high school in question boasts that their students (i.e., the population of all of their students) have an average ACT score above the national average of 18. Given your results from part D, do you believe that there is evidence that the mean ACT score at this high school is greater than 18? Explain your reasoning. p-value < 0.05 r 0.01, hence we reject the null hypothesis and conclude that there is no sufficient evidence to conclude that the mean ACT score at this high school is greater than 18