Question

4) In a survey conducted by the marketing agency 11mark, 250 of 1000 adults confessed to bringing and using their cellphone every trip to the bath room. a. Based on the result of this survey, obtain a point estimate for the proportion of adults who bring their cell phone every trip to the bathroom. b. Describe the shape of the distribution of p. C. Obtain the margin of error E for 99% confidence interval for p. d. Obtain the lower and upper bound for the 99% confidence interval for p. e. Obtain the 99% confidence interval for p and interpret it.

Answer 1

We need to construct the 99% confidence interval for the population proportion. We have been provided with the following information about the number of favorable cases:

Favorable Cases X	250
Sample Size N	1000

The sample proportion is computed as follows, based on the sample size N = 1000 and the number of favorable cases X = 250:

a) Point estimate

$\hat p = \displaystyle \frac{X}{N}$

$\hat p = \displaystyle \frac{250}{1000}$

$\hat p$ = 0.25

b) Shape of the distribution : Normal

The critical value for $\alpha$ = 0.01 is $z_c = z_{1-\alpha/2}$ = 2.576. The corresponding confidence interval is computed as shown below:

$CI(\text{Proportion}) =\displaystyle \left( \hat p - z_c \sqrt{\frac{\hat p (1-\hat p)}{n}}, \hat p + z_c \sqrt{\frac{\hat p (1-\hat p)}{n}} \right)$

$CI(\text{Proportion}) =\displaystyle \left( 0.25 - 2.576 \times \sqrt{\frac{0.25 (1- 0.25)}{1000}}, 0.25 + 2.576 \times \sqrt{\frac{0.25 (1- 0.25)}{1000}} \right)$\

Margin of error = $2.576 \times \sqrt{\frac{0.25 (1- 0.25)}{1000}}$

Margin of error = 0.035

$CI(\text{Proportion}) =\displaystyle \left( 0.25 - 0.035}, 0.25 + 0.035} \right)$

$CI(\text{Proportion}) =(0.215, 0.285)$

Therefore, based on the data provided, the 99% confidence interval for the population proportion is 0.215 < p < 0.285, which indicates that we are 99% confident that the true population proportion p is contained by the interval (0.215, 0.285).