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- A sample of n = 16 scores is obtained from a population with μ = 70 and σ = 20. If the sample mean is M = 80, what is the z -score for the sample mean?At age 9 the average weight for both boys and girls are exactly the same. A random sample of 9 year olds gave the following results.... Boys: Girls:Sample Size: 65 50Mean weight (lbs): 130 125Pop. Variance: 120 130 At alpha 0.05, do the data support the claim that there is a difference in heights?4.2 A sample of 19 female bears was measured for chest girth (y) and neck girth (x), both in inches. The least squares regression equation relating the two variables is = 5.32 + 1.53 x. The standard deviation of the regression is s = 3.692. y^ 4.2.1 What is a 95% confidence interval for the average (or mean) chest girth for bears with a neck girth of 20 inches? The standard error of the fit is s.e.(fit) = 0.880 a. (27.9, 43.9) b. (28.1, 43.7) c. (34.1, 37.8) d. None of the above 4.2.2 Interpret the confidence interval obtained in 4.2.1
- A population has µ = 60 and σ = 10. For a sample of n = 25 scores from this population, a sample mean of M = 55 would be considered an extreme value. True False3.1.1)Given X=14.5% SD=3.4% N=25 90% CI for population mean is X.where Z=1.645 at 905 confidence =14.5±1.645 =14.5±1.12 =(13.38%,15.62%) We are 90% confident that the actual mean dividend yield lies between 13.38% and 15.62%. 3.1.2) 95% CI is X±Z where Z =14.5±1.96. =14.5±1.33 =(13.17%,15.83%) is the 95% Confidence interval Compare the 2 calculations above by means of a diagram1.3.2 Consider the output given below that was obtained using the One Proportion applet. Use information from the output to find the standardized statistic for a sample propor- tion value of 0.45. Probability of success (n): 0.30 Sample size (n): 25 Number of samples: 1000 O Animate Draw Samples Total = 1000 180 Mean = 0.301 SD = 0.091 120 60 0.08 0.16 0.24 0.32 0.40 0.48 0.56 0.64 Proportion of success
- A sample of n = 64 scores has a mean of M = 68. Assuming that the population mean is p = 60, find the z-score for this sample: If it was obtained from a population with o = 16 Z = If it was obtained from a population with o = 32 Z = If it was obtained from a population with o = 48 Z =12. Two independent samples are taken from two populations. Sample 1 has a mean of 12.8, a standard deviation of 2.3, and a sample size of 18. Sample 2 has a mean of 14.2, a standard deviation of 5.2, and a sample size of 23. This gives a standard error se(Ã1 — Ă2) = 1.212. A 95% confidence interval for the difference between the two means is: (A) (-2.00, -0.80). (B) (11.66, 13.94). (C) (-3.96, 1.16). (D) (-3.85, 1.05). (E) (-3.78, 0.98).A sample of n = 16 scores is selected from a population with μ = 80 with σ = 20. On average, how much error would be expected between the sample mean and the population mean?
- The pulse rates of 176 randomly selected adult males vary from a low of 40 bpm to a high of 116 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want 95% confidence that the sample mean is within 2 bpm of the population mean. Assume that o= 10.1 bpm, based on the value s = 10.1 bpm from the sample of 176 male pulse ratesA random sample of n = 19 winter days in Denver gave a sample mean pollution index x1 = 43. Previous studies show that o1 = 10. For Englewood (a suburb of Denver), a random sample of n2 = 18 winter days gave a sample mean pollution index of x2 = 34. Previous studies show that o2 = 13. Assume the pollution index is normally distributed in both Englewood and Denver. Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a 1% level of significance. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H1 H2 O Ho: H1 = l2; H1: H1 < µ2 (b) What sampling distribution will you use? What assumptions are you making? O The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. O The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. O The Student's…A random sample of 130 observations produced a mean of I = 23.5 and a standard deviation s = 2.16. (a) Find a 90% confidence interval for u (b) Find a 95% confidence interval for u (c) Find a 99% confidence interval for u