Let X be an arbitrary random variable with mean and standard deviation o. Let X be its sample distribution for n = 100 sample size. 1. What is the expected value of the sample distribution in terms of μ? 2. What is the standard error of the sample distribution in terms of o?
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- Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with ? = 3.2%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 The sample mean is x = 5.38%. Suppose that for the entire stock market, the mean dividend yield is ? = 4.3%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.3%? Use ? = 0.01. (a) What is the level of significance?State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H0: ? > 4.3%; H1: ? = 4.3%; right-tailedH0: ? = 4.3%; H1: ? ≠ 4.3%; two-tailed H0: ? = 4.3%; H1: ? < 4.3%; left-tailedH0: ? = 4.3%; H1: ? > 4.3%; right-tailed (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The Student's t, since n is large with unknown ?.The standard normal, since we assume that x has a normal…Production output (i.e., number of parts) for a random sample of days from two different plants is shown below. Sample Size (n) 45 55 Sample Mean output (x-bar) 1,250 1,325 Sample Variance (s2) 625 676 What is the estimate of the standard deviation for the difference between the two means?Let X be a normal random variable with mean 48 and standard deviation 3.8. Suppose that x = a has a z-score of z = 2.5. How do we interpret this z-score? Group of answer choices: x = a has a difference of 2.5 with x = 48 x=a is 2.5 standard units away from x = 48. x = a is 3.8 standard units away from x = 48 x = a is has a difference of 3.8 from x = 48.
- A population of values has a normal distribution with M=153.9 and O=82.9 You intend to draw a random sample of size n=115. A). Find P86, which is the score separating the bottom 86% scores from the top 14% scores. P86 (for single values) =__________ B). Find P86, which is the mean separating the bottom 86% means from the top 14% means. P86 (for sample means) =_________Assume the random variable X is normally distributed, with mean μ=41 and standard deviation σ=7. Find the 14th percentile. The 14th percentile is nothing.How is the t distribution different from the Z distribution? a. The t distribution does not have any negative numbers. b. The t distribution is asymmetrical; that is, it is not a mirror image on each side of the population mean. c. The t distribution is wider than the Z distribution due to lower sample size. d. All of the above are differences between the t and Z distributions
- Can a normal approximation be used for a sampling distribution of sample means from a population with u = 55 and o = 10, when n = 4? Answer E Tables E Keypad Keyboard Shortcuts O Yes, because the mean is greater than 30. O Yes, because the sample size is less than 30. O No, because the sample size is less than 30. O No, because the standard deviation is too small.Let x be a random variable representing dividend yield of bank stocks. We may assume that x has a normal distribution with ? = 2.5%. A random sample of 10 bank stocks gave the following yields (in percents). 5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1 The sample mean is = 5.38%. Suppose that for the entire stock market, the mean dividend yield is ? = 4.9%. Do these data indicate that the dividend yield of all bank stocks is higher than 4.9%? Use ? = 0.01. (c) Find (or estimate) the P-value. (Enter a number. Round your answer to four decimal places.) ?Sketch the sampling distribution and show the area corresponding to the P-value. (Select the correct graph.)Independent simple random samples are taken to test the difference between the means of two populations whose variances are not known. The sample sizes are n1 = 12 and n2 = 14. Which is the correct distribution to use?
- The distribution of average hours worked per week for random samples of 1000 workers in the U.S. has a mean of 40 and a standard error of 0.04 The distribution in this scenario is a: sampling distribution sample distribution population distributionA population has a mean μ=40 and a standard deviation σ=4. What is the standard deviation of the sampling distribution of the sample means if the sample size is n=16?A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. State the conclusion for the test. Use a 0.01 significance level to test the claim that men have a higher mean body temperature than women. μ n X S Men H₁ 11 97.66°F 0.75°F Women H₂ 59 97.22°F 0.68°F O A. Reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women. O B. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women. O C. Reject the null hypothesis. There is sufficient evidence to support the claim that men have a higher mean body temperature than women. O D. Fail to reject the null hypothesis. There is sufficient evidence to support the…