Question

13. A police department head wants to estimate the average hight of his crew. To do this estimation with level of confidence 98%, a sample of 50 members was taken and resulted a sample mean of 178 cm and a sample standard deviation of 50 cm. (i) The critical value equals A. 2.0086 kg B. 2.4049 kg C. 2.0096 kg D. 2.4033 kg (ii) The confidence interval is A. [163.790, 192.210] B. [160.995, 195.005] C. [161.006, 194.994] D. [163.797, 192.203) (iii) The confidence interval is interpreted as follows A. The department head is 98% confident that the average height of his crew is different from 178 cm B. The department head is 98% confident that the average height of his crew is inside the confidence interval. C. The department head is 98% confident that the average height of his crew is outside the confidence interval. D. The department head is 98% confident that the sample mean height of his crew is inside the confidence interval.

Answer 1

Given that: n=50; x =178 and s=50 i) Calculate the critical value. From the t- tabulated values, the t - critical value at the 98% confidence level with 50-1= 49 degrees of freedom is: 2.4049 ii) Calculate the 98% confidence interval. S 98%C1 = t critical 50 = 178 +2.4049 V50 = 178 +17.005 =(160.995,195.005) iii) The department head 98% confident that the average height of his crew is inside the confidence interval