A random sample of 40 transmission replacement costs finds the mean to be $2640.

Question

A random sample of 40 transmission replacement costs finds the mean to be $2640.00. Assume the population standard deviation is $425.00 A 90% confidence interval for the population mean repair cost is (2529.5, 2750.5). Change the sample size to n = 80. Construct a 90% confidence interval for the population mean repair cost. Which is confidence interval wider? Explain. Construct a 90% confidence interval for the population mean repair cost. the 90% confidence interval is (Round to one decimal place as needed.)

Explanation / Answer

a. Here n=40 and population sd is known so we will take z distribution.

For 90% CI z=1.645, as from z table we can see that P(-1.645<z<1.645)=0.90

Now sd=425 so Margin of Error=E=z*sd/sqrt(n)=1.645*425/srt(40)=110.541

Hence CI=mean+/-E=2640+/-110.541=(2529.5,2750.5)

Similarly for n=80 E=1.645*425/sqrt(80)=78.165

So CI=2640+/-78.165=(2561.8,2718.2)

CI with n=40 is wider

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