Question

A The housing market has recovered slowly from the economic crisis of 2008. Recently, in one large community, realtors randomly sampled 41 bids from potential buyers to estimate the average loss in home value. The sample showed the average loss was $9424 with a standard deviation of $1269 Find a 95% confidence interval for the mean loss in value per home. ($ $ ) B. In a national health survey, the total cholesterol of 2827 adults in a particular country averaged 188.6 mgdL with a standard deviation of 41.99 mg/dL Sketch and clearly label the sampling model of the mean cholesterol levels of samples of size 2827 based on the 68-95-99.7 Rule 68% 95% 99.7% C. Data on the fuel economy of several 2010 model vehicles are given in the accompanying table. mpg 33 Ovana 32 35 13 26 14 39 15 30 Find and interpreta 95% confidence interval for the gas mileage of 2010 vehicles. One is 95% confident that the true mean gas mileage for cars like the ones in the sample is between mpg and mpg. D. A company announced a "1000 Chips Trial" claiming that every 18-ounce bag of its cookies contained at least 1000 chocolate chips. Students purchased random bags of cookies from different stores and counted the number of chips in each bag. 1048 1101 1187 1203 1263 1234 1222 1356 1391 1443 Create a 95% confidence interval for the average number of chips. .

Answer 1

A)

n = sample size = 41

Sample mean = $\bar{x}=9424$

Sample standard deviation = s = 1269

Here population standard deviation is not known but sample size n is large (n > 30)

So we use z interval.

95% Confidence interval for population mean is

$\bar{x}-z_c*\frac{s}{\sqrt{n}}<\mu<\bar{x}+z_c*\frac{s}{\sqrt{n}}$

where zc is z critical value for (1+c)/2 = (1+0.95)/2 = 0.975 is

zc = 1.96 (From statistical table of z values)

1269 1269 9424 -1.96*4 行くり<9424 +1.96* V41

$9424-1.96*198.1845<\mu<9424+1.96*198.1845$

$9424-388.4416<\mu<9424+388.4416$

$9035.558<\mu<9812.442$

95% Confidence interval for the mean loss in value per home is ($9035.558, $9812.442)