MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- 10 squirrels were found to have an average weight of 610 grams with a sample standard deviation of 7.75 . Find the 95 % confidence interval of the true mean weight (assume the t-student distribution). A.)(608.25, 614.16) B.)(561.02, 614.16) C.) (604.46,615.54) D.) (513.79, 612.77)arrow_forwardTwo samples were given the same task. One sample was put into a room painted red, and the pther was put into a blue room. Thier creativity was scored and recorded while completeing the task. Results are given below. Use a 0.05 significance level to test the claim that the samples from populations with the same standard deviation. Assume that all the measurements are met for this test. Red Room n=34 Xbar =15.92 s= 5.97 Blue Room n=36 Xbar= 12.95 s= 5.49 Is this a right, left or two-tailed test? What is the P-Value? Rounded to 3 decimals Write a conclusion about this test.arrow_forwardFrom a sample of 38graduate students, the mean number of months of work experience prior to entering an MBA program was 34.93. The national standard deviation is known to be 18 months. What is a 95% confidence interval for the population mean? 95% confidence interval for the population mean is _____ , _____. (Use ascending order. Round to two decimal places as needed.)arrow_forward
- A researcher is attempting to calculate a 95% confidence interval for the mean average wingspan of Statsian Swallows in a particular rainforest. They take a sample and find that of the 16 swallows surveyed, the mean average was 3.14 inches and the standard deviation was 0.2 inches. What is the 95% confidence interval for this sample?arrow_forwardIn a random sample of twelve people, the mean driving distance to work was 18.8 miles and the standard deviation was 4.9 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean μ. Interpret the results.arrow_forwardA simple random sample of 5 observations from a normal distribution was taken, and the following values were obtained: {15, 19, 19, 20, 22}. A 95% confidence interval for the mean is O A. (16.569, 21.431) O B. (15.835, 22.165) O C.(16.069, 21.931) O D. (17.124, 20.876) O E. (16.765, 21.235)arrow_forward
- 300 high school students were asked how many hours.of TV they watch per day. The mean was 2 hours, with a standard deviation of .5. Using a 90% confidence level, calculate the maximum error of estimate A.7.43% B.0.27% C.5.66% D.4.75%arrow_forwardThe average income for a county in Texas is reported in a study. The margin of error is 767.42. Use a confidence interval of 95%. What is the approximate standard deviation of the sample assuming n = 45? O 4,598.04 O 8,315.63 O 2,574.36 O 2,626.54arrow_forwardIn a random sample of 50 people, the mean body mass index (BMI) was 27.7 and the standard deviation (s) was 6.12. a. Construct a 90% confidence interval for the population mean. b. Construct a 96% confidence interval for the population mean.arrow_forward
- From a sample of 37 graduate students, the mean number of months of work experience prior to entering an MBA program was 36.63. The national standard deviation is known to be 19 months. What is a 90% confidence interval for the population mean? A 90% confidence interval for the population mean is 10 (Use ascending order. Round to two decimal places as needed.) ...arrow_forward900 students were surveyed and had an average GPA of 2.7 with a standard deviation of 0.4. Calculate the margin of error for a 90% confidence level:arrow_forwardSuppose the population standard deviation is = 5, an SRS of n = 100 is obtained, and the confidence level is chosen to be 99.5%. The margin of error for estimating a mean is given by: a. 1.576. b. 1.165. c. 1.404. d. 1.223.arrow_forward
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