Construct a 95% confidence interval for the population mean, based the 2, 3, 4,

Question

Construct a 95% confidence interval for the population mean, based the 2, 3, 4, 5, 6, and 20 Change the number20 to 7 and recalculate the confidence interval. Using these results, describe the effect of an outlier i.e., extreme value) on the confidence interval. Applying the Concepts You can solve Problems 8.15-8.22 with or without Microsoft Excel, Minitab, or SPSS. You should use Microsoft Excel, Minitab, or SPSS to solve problem 8. 23.__ 8.15 The owner of a stationery store wants to estimate the mean retail value of greeting cards that the store has in its inventory. A random sample of 20 greeting cards indicates mean value of $1.67 and a standard deviation of $0.32. a. Assuming a normal distribution, construct a 95% confidence interval estimate of the mean value of all greeting cards in the store's inventory. b. How are the results in (a) useful in assisting the store owner to estimate the total value of her inventory?

Explanation / Answer

Given that N = 20 m mean = 1.67 and SD = 0.32

now we know that the confidence interval is given as

CI = Mean +- Z*SD/sqrt(N) , now Z = 1.96 for the 95% confidence interval (using the z table, please keep them handy)

1.67 +- 1.96*0.32/sqrt(20)

= (1.529, 1.810)

b)

This means that the true mean of the population would lie between 1.529 and 1.810 dollars.

Now as we have the range of true mean in place, then we can simply multiply the total number of units in the population inventory to arrive at the range of total value of the inventory. we can do all this with 95% confidence

so if the store has 100 units in total

then then inventory value is between 152.9 and 181 dollars

Hope this helps

  

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