Question

sxS (Round to three decimal places as needed ) b Construct a 90% contence interval estrate ofthe po uta onr ortion tofonine shoppers who would be inuenced to spend more money (ine weh dsorts of red while shoppin sxsDRourd to three decrnal places as needed ) %confidence nerval est c.cnstruct a 9 ate of the po ulation roportio or on est o rs who woud tenn ned to spend more money on. een ract S(Round to three decimal places as needed ) For part a) the sample size is For part b). the sample size is For part c. the sample size is O Type here to search

Answer 1

Solution: a. Construct a 90% confidence interval estimate of the population proportion % of online shoppers who would be influenced to spend more money online with free shipping.

Answer: The 90% confidence interval estimate of the population proportion is:

$\hat{p} \pm z_{\frac{0.1}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

$\hat{p}=0.32$ is the sample proportion of those who said free shiping

$z_{\frac{0.10}{2}}$ is the critical value at 0.10 significance level and is given:

$z_{\frac{0.10}{2}}=1.645$

$n=953$

0.321.645.210.32) 953

  $0.32\pm 0.0249$

  $\left [ 0.32-0.0249,0.32+0.0249 \right ]$

$\left [ 0.295,0.345 \right ]$

Therefore, 90% confidence interval is:

$0.295\leq$ % $\leq 0.345$

b.Construct a 90% confidence interval estimate of the population proportion % of online shoppers who would be influenced to spend more money online with discounts offered while shopping

Answer: The 90% confidence interval estimate of the population proportion is:

$\hat{p} \pm z_{\frac{0.1}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

$\hat{p}=0.22$ is the sample proportion of those who said offering discounts while shopping

$z_{\frac{0.10}{2}}$ is the critical value at 0.10 significance level and is given:

$z_{\frac{0.10}{2}}=1.645$

$n=953$

$\therefore 0.22\pm 1.645\sqrt{\frac{0.22(1-0.22)}{953}}$

  $0.22\pm 0.0221$

  $\left [ 0.22-0.0221,0.22+0.0221 \right ]$

$\left [ 0.198,0.242 \right ]$

Therefore, 90% confidence interval is:

$0.198\leq$ % $\leq 0.242$

c.Construct a 90% confidence interval estimate of the population proportion % of online shoppers who would be influenced to spend more money online with product review

Answer: The 90% confidence interval estimate of the population proportion is:

$\hat{p} \pm z_{\frac{0.1}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

$\hat{p}=0.14$ is the sample proportion of those who said product review

$z_{\frac{0.10}{2}}$ is the critical value at 0.10 significance level and is given:

$z_{\frac{0.10}{2}}=1.645$

$n=953$

$\therefore 0.14\pm 1.645\sqrt{\frac{0.14(1-0.14)}{953}}$

  $0.14\pm 0.0185$

  $\left [ 0.14-0.0185,0.14+0.0185 \right ]$

$\left [ 0.122,0.158 \right ]$

Therefore, 90% confidence interval is:

$0.122\leq$ % $\leq 0.158$

d. You have been asked to update the results of this study. Determine the sample size necessary to estimate, with 90% confidence, the population proportion in (a) through (c) to within $\pm 0.04$

Answer: For part a, the sample size is:

$n=\hat{p}(1-\hat{p})\left ( \frac{z_{\frac{0.1}{2}}}{0.04} \right )^{2}$

$=0.32(1-0.32)\left ( \frac{1.645}{0.04} \right )^{2}$

$=368.02\approx 369$

For part b, the sample size is:

$n=\hat{p}(1-\hat{p})\left ( \frac{z_{\frac{0.1}{2}}}{0.04} \right )^{2}$

$=0.22(1-0.22)\left ( \frac{1.645}{0.04} \right )^{2}$

$=290.22\approx 291$

For part c, the sample size is:

$n=\hat{p}(1-\hat{p})\left ( \frac{z_{\frac{0.1}{2}}}{0.04} \right )^{2}$

$=0.14(1-0.14)\left ( \frac{1.645}{0.04} \right )^{2}$

$=203.63\approx 204$