Answer:
The answer is "23, 62, and 139"
Step-by-step explanation:
In point a:
[tex]\to \sigma = \$0.24\\\\ \to E = \$0.10\\\\ 95\% \text{confidence level of the z}: \\\\\to \alpha = 1 - 95\% = 1 - 0.95 = 0.05\\\\\to \frac{\alpha}{2} = \frac{0.05}{2} = 0.025\\\\\to Z_{\frac{\alpha}{2}} = Z_{0.025} = 1.96\\\\[/tex]
Calculating the sample size:
[tex]\to n = (\frac{(Z_{\frac{\alpha}{2}} \times \sigma)}{E})^2[/tex]
[tex]= (\frac{(1.96 \times 0.24 )}{0.10})^2\\\\= 22.13 \approx 23[/tex]
In point b:
[tex]\to \sigma = \$0.24\\\\\to E = \$0.06\\\\\to Z_{\frac{\alpha}{2}} = Z_{0.025} = 1.96\\\\\to n = (\frac{(Z_{\frac{\alpha}{2}} \times \sigma )}{E})^2\\\\[/tex]
[tex]= (\frac{(1.96 \times 0.24 )}{0.06})^2\\\\= 61.47 \approx 62[/tex]
In point c:
[tex]\to \sigma = \$0.24\\\\\to E =\$0.04\\\\\to Z_{\frac{\alpha}{2}} = Z_{0.025} = 1.96\\\\\to n = (\frac{(Z_{\frac{\alpha}{2}} \times \sigma )}{E})^2\\\\[/tex]
[tex]= (\frac{(1.96\times 0.24 )}{0.04})^2 \\\\=138.30 \approx 139[/tex]
