Compute the confidence intervals for the following scenarios: #1-The inspection

Question

Compute the confidence intervals for the following scenarios: #1-The inspection division of the Lee County Weights and Measures Department is interested in estimating the actual amount of soft drink that is placed in 2- liter bottles at the local bottling plant of a large nationally known soft-drink company. The bottling plant has informed the inspection division that the standard deviation for 2-liter bottles is 0.05 liter. A random sample of one hundred 2-liter bottles obtained from this bottling plant indicates a sample average of 1.99 liters. State a 95% confidence interval for the mean amount of soft drink that is placed in 2-liter bottles. % confident that the amount of soft drink that is placed in 2-liter botties at this plant is between and liters. #2 . A stationary store wants to estimate the average retail value of greeting cards that it has in its mentory. Knowing that the retail value varies normally with a standard deviation of $0.32. A random sample of 20 cards produced the following prices: 2.42 .99 197 3.64 1.45 2.50 3.10 .991.50 2.25 1 3.60 4.50 .99 2.25 3.60 State a 90% confidence interval for the mean retail value of the greeting cards. % confident that the retail value of the greeting cards is between $ and $

Explanation / Answer

1)

I am 95% confident that the mean amount of soft drink that is placed in 2 liter bottles at this plant is between 1.9802 and 1.9998

2)

(2.2628 , 2.5102)

I am 90% confident that the average retail value of greeting cards is between 2.2628 and 2.5102

CI for Mean CI for 95% n 100 mean 1.99 z-value of 95% CI 1.9600 std. dev. 0.05 SE = std.dev./sqrt(n) 0.00500 ME = z*SE 0.00980 Lower Limit = Mean - ME 1.98020 Upper Limit = Mean + ME 1.99980 95% CI (1.9802 , 1.9998 )

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