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1) A random sample of size 100 was taken from a population. A 94% confidence interval to estimate the mean of the population was computed based on the sample data. The confidence interval for the mean is: (107.62,129.75). What is the z-value that was used in the computation. Round your z-value to 2 decimal places.

To answer the problem input only the actual number. Do not include units. Do not give your answer in sentence form -- just include the numerical answer rounded to exactly 2 decimal places.

2) The U.S. Food and Drug Administration (FDA) performed a study that compared the carbohydrate content (as a percentage of weight) of several major brands of corn and potato chips. Fourteen brands of corn chips were sampled and 14 brands of potato chips were sampled. Data was collected and the following 95% confidence interval was computed for the difference in means. Intrepret the 95% confidence interval for the difference in the mean carbohydrate content between the Corn Chip and Potato Chip groups.

The C.I. is: (27.292,39.836)

where {mu}_1 = the average carbohydrate content as a percentage of weight in corn chips.

and {mu}_2 = the average carbohydrate content as a percentage of weight in potato chips.

Select the best answer below.

A. We are 95% confident that there is no statistically significant difference in the average carbohydrate content as a percentage of weight between corn chips and potato chips.

B. We are 95% confident that the average carbohydrate content as a percentage of weight in corn chips is lower than the average carbohydrate content as a percentage of weight in potato chips.

C. We are 95% confident that the average carbohydrate content as a percentage of weight in corn chips is higher than the average carbohydrate content as a percentage of weight in potato chips.

D. None of these are correct.

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  • Category:- Statistics and Probability
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