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- The circumference of a particular species of tree was thought to be different depending on certain geographical factors. An SRS of 60 trees is selected in Sequoyah National Forest, where it is found that = 6 meters with a standard deviation = 3 meters. A second SRS of size 40 is independently selected in Yellowstone National Forest, and it is found that = 4 meters with a standard deviation = 2 meters. Let μ1 and μ2 represent, respectively, the true mean circumferences of this particular species of tree. Assume the standard deviations of the two populations are equal. The researcher tests the following hypotheses: H0 : μ1 = μ2, Ha : μ1 ≠ μ2The numerical value of the two-sample t statistic isThe circumference of a particular species of tree was thought to be different depending on certain geographical factors. An SRS of 60 trees is selected in Sequoyah National Forest, where it is found that = 6 meters with a standard deviation = 3 meters. A second SRS of size 40 is independently selected in Yellowstone National Forest, and it is found that = 4 meters with a standard deviation = 2 meters. Let μ1 and μ2 represent, respectively, the true mean circumferences of this particular species of tree. Assume the standard deviations of the two populations are equal. A 95% confidence interval for μ1 – μ2 isGiven a population with a mean 176 and standard deviation, σ. (Case A) Repeatedly take samples of size 28 from this population and in each case compute the sample mean, . Let x-bar(A)and sigma(A) be respectively the mean and the standard deviation of these sample means. (Case B) Repeatedly take sample of size 64 from this population and in each case compute the sample mean, . Let x-bar(B) and sigma(B) be respectively the mean and the standard deviation of these sample means. If the population standard deviation, σ, is 32, then 95% of the sample means in case B what are the two intervals?
- Suppose we want to test whether or not the average SAT score at the University of Seychelles is significantly larger than the average SAT score at Vermont. Taking a random sample, we manage to get the following seven scores from Vermont students: Vermont: 1340 1500 1430 1440 1380 1470 1290 and the following nine scores from Seychelles students: Seychelles: 1540 1480 1390 1450 1440 1350 1520 1400 1600 Assuming that the standard deviation in SAT scores for students at Seychelles and Vermont is the same and equal to 77 points, test at the α=:05 significance level whether or not the average SAT score atthe University of Seychelles is larger than the average SAT score at Vermont.2. The values of the account receivable for a department store are normally distributed with a population mean of $260.00 and standard deviation of $15.00. (a) What proportion of accounts receivable are of values between $248.00 and $268.00? (b) If an auditor takes many samples of size 16 from the population of accounts receivable: (i) What will be the mean and the standard deviation of the sample means? (ii) What is amount will be exceeded by 67% of the sample means? (iv) What is the 80% confidence intervalof the mean?1. Suppose that adult glucose levels are Normally distributed with a mean of 99 mg/dL (milligrams per decilitre) and a standard deviation of 16 mg/dL. We take an SRS of 145 adults, measure their glucose levels, and calculate the sample mean. (a) If we were to take many such samples, what would be the approximate sampling distribution of the resulting sample means? Show your work, by hand. (b) Hyperglycemia is often classified as a glucose level greater than 125 mg/dL. What proportion of individuals in a single sample do we expect to have hyperglycemia? (Use R, or calculate by hand, show all work) (c) What is the probability that the mean of a sample of size 145 is greater than 125 mg/dL? (Use R, or calculate by hand, show all work) (d) Between which values would the middle 60% of sample means lie? (Use R, or calculate by hand, show all work)
- If a population has standard deviation 20, what is the minimum sample size to be 95% confident that the error should be accurate to within 4?1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. At 5% level, what are the critical values for testing equality of mean weights in problem 1? A. 2.18 B. -2.18 and 2.18 C. -1.78 D.-1.78 and 1.78 3.What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical…1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2.What would be the degree of freedom for the test statistic in problem 1? A. 6 B. 9 C. 12.7 D. 14 3. What would be the computed test statistic in problem 1? A. 2.93 B. 3.57 C. 8.44 D. 11.48
- 1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C. H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D.The computed…1. The sample mean weights for two varieties of lettuce grown for 16 days in a controlled environment are 3.259 and 1.413 and the corresponding sample standard deviations are .400 and .220. If the sample sizes for the two varieties are 9 and 6 respectively, what would be the pair of hypotheses to test if the two varieties of lettuce have the same average weight? (Given: weight of each variety of lettuce is normally distributed). A. H0: μ1 ≠ μ2 vs H1: μ1 = μ2 B. H0: μ1 = μ2 vs H1: μ1 ≠ μ2 C.H0: μ1 > μ2 vs H1: μ1 ≤ μ2 D. H0: μ1 ≤ μ2 vs H1: μ1 > μ2 2. What is the best decision using critical value approach in problem 1? A. The computed test statistic falls in the critical region and we do not reject the null hypothesis. B. The computed test statistic does not fall in the critical region and we do not reject the null hypothesis. C. The computed test statistic falls in the critical region and we reject the null hypothesis. D. The computed test statistic does not fall…1)Suppose that the weights of 3200 registered female Great Danes in the United States are distributed normally with a mean of 115 lb. and a standard deviation of 5.4 lb.Approximately how many of the Great Danes weigh more than 125.8 lbs.? Round to the nearest whole number 2) Body mass index, or BMI, of 20-year-old men is normally distributed with a mean of 23 and a standard deviation of 3.8. Twenty-year-old Allan’s BMI has az-score of 1.25. What is Allan’s BMI? 3) An experiment compares weight of fish for two different brands fish food. Nacho Average Minosfish food has a mean fish weight 10 lbs. and a standard deviation of 1.5 lbs. Impossible Minos fish food has a mean fish weight of 12 lbs. and a standard deviation of 1 lb. A fish is measured to be 11.25 lbs. Which brand of fish food, Nacho Average Minos or Impossible Minos, is more likely to have been used to feed this fish? 4) Height of 10th grade boys is normally distributed with a mean of 63.5 in. and a standard deviation of 2.9…