Question

Problem: A marketing researcher randomly selects 20 female students in a certain university and found out that their

average monthly expenditures for their cell phone loads is Php 500.00 with a standard deviation of Php75.00

a. Find the 99% confidence interval for the average expenditures of female students for their cell phone loads.

b. Find the 90% confidence interval for the average expenditures of female students for their cell phone loads.

c. Find the 80% confidence interval for the average expenditures of female students for their cell phone loads.

Answer 1

Confidence interval can be found the following way:

$x \in (a-Cr_{\alpha}*{\frac {\sigma} {\sqrt n}};a+Cr_{\alpha}*{\frac {\sigma} {\sqrt n}})$ , where a-sample mean, $Cr_{\alpha}$ - critical value on ${\alpha}$ confidence level, $\sigma$ - standard deviation, n - sample size

Since the sample size is small(<30), then it is appropriate to use t-score as critical value. So, $P(T(n-1)>Cr)={\frac {1+\alpha} 2}$ , where T(n-1)-Student's distribution with n-1 degrees of freedom

Which means, the sought confidence interval is $(500-Cr_{\alpha}*{\frac {75} {\sqrt {20}}};a+Cr_{\alpha}*{\frac {75} {\sqrt {20}}})=(500-16.77*Cr_{\alpha},500+16.77*Cr_{\alpha})$

a) $P(T(19)>Cr)={\frac {1+0.99} 2}=0.995\implies Cr=2.861$

So, the sought confidence interval is $(500-16.77*2.861,500+16.77*2.861)=(452, 548)$

b) $P(T(19)>Cr)={\frac {1+0.9} 2}=0.95\implies Cr=1.729$

So, the sought confidence interval is $(500-16.77*1.729,500+16.77*1.729)=(471, 529)$

c) $P(T(19)>Cr)={\frac {1+0.8} 2}=0.9\implies Cr=1.328$

So, the sought confidence interval is $(500-16.77*1.328,500+16.77*1.328)=(478, 522)$