Question

13. 4 GB Flash Memory Card After scanni of a 4 GB flash memory card using his smartp emory Card After scannine the UPC code Sullivan received the followineries for the goods at of online retailers Treat the data as simple random of all online retailers. 12.25 13.49 13.76 14.75 14.99 20.49 15.27 1595 17.88 Source: Scan Life his smartphone, Michael owing prices for the goods at a variety lata as a simple random sample 14.99 (a) Determine a point estimate for the population mean price ol the memory card. (b) Use the normal probability plot and boxplot shown next to assist in verifying the requirements for constructing a confidence interval for the mean price. 12 LLLLLL 14 16 18 20 Price (c) Construct and interpret a 95% confidence interval to estimate the mean price of a 4 GB flash memory card. (d) Construct and interpret a 99% confidence interval to estimate the mean price of a 4 GB flash memory card (e) What happens to the interval as the level of confidence is increased? Explain why this is a logical resulta . . Healthcare A Gallup poll conducted in November 2010 found that 493 of 1050 adult Americans believe it is the responsibility of the federal government to make sure all Americans have healthcare coverage. (a) Obtain a point estimate for the proportion of adult Americans who believe it is the responsibility of the federal government to make sure all Americans have healthcare coverage. (b) Verify that the requirements for constructing a confidence interval for p are satisfied. (e) Construct a 95% confidence interval for the proportion of adult Americans who believe it is the responsibility of the federal government to make sure all Americans have healthcare coverage (d) Construct a 99% confidence interval for the proportion of adult Americans who believe it is the responsibility of the federal government to make sure all Americans have healthcare coverage (e) What is the effect of increasing the level of confidence on the width of the interval?

Answer 1

Part a)

We have to find Point estimate x̅

∑x = 153.82

n = 10

Mean , x̅ = Ʃx/n = 153.82/10 = 15.3820

Point estimate, x̅ = 15.3820

b) Normal probability plot and boxplot that shows distribution is symmetric. so we consider distribution is approximately normal. We use t test for construction confidence interval.

c) 95% Confidence interval :  

At α = 0.05 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.05, 9) = 2.262

Lower Bound = x̅ - t-crit*s/√n = 15.382 - 2.262 * 2.3427/√10 = 13.71

Upper Bound = x̅ + t-crit*s/√n = 15.382 + 2.262 * 2.3427/√10 = 17.06

d) 99% Confidence interval :  

At α = 0.01 and df = n-1 = 9, two tailed critical value, t-crit = T.INV.2T(0.01, 9) = 3.250

Lower Bound = x̅ - t-crit*s/√n = 15.382 - 3.25 * 2.3427/√10 = 12.98

Upper Bound = x̅ + t-crit*s/√n = 15.382 + 3.25 * 2.3427/√10 = 17.79

You can verify Confiidence interval answer using Minitab :

One-Sample T: Data Variable Data N 10 Mean St Dev 15.385 2.340 SE Mean 0.740 95% CI (13.711, 17.059) One-Sample T: Data Variable Data N 10 Mean StDev SE Mean 15.385 2.340 0.740 99% CI (12.980, 17.790)

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