Question

In a random sample of six mobile​ devices, the mean repair cost was ​$80.00 and the standard deviation was ​$12.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean mu is

Answer 1

Solution: The provided Yo formation is sis follasuks Saraple Bean -80 Sample size sample standard devisition s= 12 confidence level lange 0.95 To find The mangin of enor and construct a Confidence internal for the population mean. 95% The cumber of degrees of freedom ore dfiola 6-135i and the significance level is dö 0105 Based on the provided information, the critical t-value for a =0.05 and df = 5 degrees of freedom to = 2.57) The margin of enor for the population meani; Maoglio of prox E = 1 ts xs sh f. 2.57)_412 I 12.593 Margin of prror {=. I 12. 593 The 95% confidence for the is como puted as follows population mean لي CI f - to xs Vi X + texs too 80 - 2.571*12 80 f 2.57 x 12 va

= (80-12:593, 80+ 120593) CI = (67-403, 92.593) Interpretation Therefore based on the data provided the 95% confidence internal for population mean l is 67.407 <lc 92.593 which indicates that we are 95% confident that the mean repair cost is contained by the interval ( 67.407, 92, 593)