Question

Please help me solve this and how do I decide which confidence level I should use z or t?

Try yourself 1 ► The owner of a small tailoring shop keeps careful records of all alterations. Two common alterations to men's clothing are lengthening the inseam on pants and letting out a sports coat around the waist. A random sample of each type of alteration was obtained, and the summary statistics are given in the table below. Measurements are in inches. Assumed Sample Sample size Alteration mean o 0.74 0.22 Pants inseam Sports coat waist 33 42 1.05 0.37 a) Find a 95% confidence interval for the mean alteration of the inseams on men's pants. b) How large a sample is necessary for the bound on the error of estimation to be 0.05 for the confidence interval in part (a)? c) Find a 90% confidence interval for the mean alteration of the waists of sports coats. d) How large a sample is necessary for the bound on the error of estimation to be 0.07 for the confidence interval in part (c)?

Answer 1

Sol:

a)

we want to find the 95% confidence interval for the mean alteration of the inseams on men's pants.

$\bar{X1}\pm Z_{\alpha/2}*\sigma_1/\sqrt{n1}$

$0.74\pm Z_{0.05/2}*0.22/\sqrt{33}$

$0.74\pm 1.96*0.0383$ ; Z- value from std. normal table at alpha=0.05

$0.74\pm0.0751$

$[0.66,0.82]$

The sample mean are lies between 0.66 and 0.82.

b)

Therefore,

E=0.05

The sample size n is

$n=(\frac{Z_{\alpha/2}\sigma_1}{E})^2$

$=(\frac{Z_{0.05/2}*0.22}{0.05})^2$

  $=(\frac{1.96*0.22}{0.05})^2$ ; Z- value from std. normal table at alpha=0.05

  $=74$

The 74 large a sample is necessary foe 95 % confidence interval .

c)

we want to find the 90% confidence interval for the mean alteration of the waists of sports coasts.

$\bar{X2}\pm Z_{\alpha/2}*\sigma_2/\sqrt{n2}$

$1.05\pm Z_{0.10/2}*0.37/\sqrt{42}$

$1.05\pm 1.64*0.0571$ ; Z- value from std. normal table at alpha=0.10

$1.05\pm0.0936$

$[0.96,1.14]$

The sample mean are lies between 0.96 and 1.14.

d).

Therefore,

E=0.07

The sample size n is

$n=(\frac{Z_{\alpha/2}\sigma_2}{E})^2$

$=(\frac{Z_{0.10/2}*0.37}{0.07})^2$

  $=(\frac{1.64*0.37}{0.07})^2$ ; Z- value from std. normal table at alpha=0.10

  $=75$

The 75 large a sample is necessary foe 90 % confidence interval .

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