Question

please answer all parts in detail :)

uppose the weights of parts in one lot has a normal distribution. Six parts were randomly drawn from this lot and weighed. Their weights were 21.6 oz. 22.4 oz. 21.4 oz. 20.2 oz. 22.0 oz 20.8 oz (a) (2 points] Compute the sample standard deviations of these six parts. (b) [4 points] Construct a 99% two-sided confidence interval for the true mean weight of all parts in this lot. (c) [4 points] Compute a 99% confidence upper bound for the variance of the weights of all parts in this lot.

Answer 1

Thursday

30/07/2020

From given data = A) 21.6 22.4 21.4 20.2 22 20.8 B) Given that X = 21.4 S = 0.8 Therefore, a/2 = 0.005 Therefore, the 99% confidence interval of the population mean is = c) Given that n = 6 S= 0.8 CI = 99 n-1= 5 Therefore, the 99% confidence interval for population variance is between = na 6 Here, Now, finding mean and SD = CI = 99 df = 6-1 Lower limit= = X-E = 21.4-1.317 = 20.083 x = 21.4 tca = to.005,5 = 4.032 = 5 At 99 % confidence level = Upper bonda = (n-1)s-2/X^2L = 5*0.842 /0.412 = 3.2 /0.412 = 7.77 Sa 0.80 (A) Now, Standard errors = s/n = 0.8/16 = 0.3266 Upper limita = X+E = 21,4+1.317 = 22.717 Margin of error = ta/2.,df std error = 4.032*0.3266 - 13168 a = 1-9995 = 1-0.99 = 0.01 Therefore, a/2 = 0.005 & Za /2 = 2.58 At 99 % confidence leveltis = 1-9996 = 1-0.99 = 0.01 Now lets calculate, Critical value of chi square distribution a) Lower critical value(X^2L) = 0.412 b) Upper critical valueſX^2U) - 16.75 {by using chi square critical value table}

pls upvote...!!!