Question

A survey of 270 young professionals found that one dash eighth of them use their cell phones primarily for​ e-mail. Can you conclude statistically that the population proportion who use cell phones primarily for​ e-mail is less than 0.17​? Use a​ 95% confidence interval. The​ 95% confidence interval is [nothing comma nothing ]. As 0.17 is ▼ within the limits below the lower limit above the upper limit of the confidence​ interval, we ▼ cannot can conclude that the population proportion is less than 0.17.

Answer 1

Answer:

As 0.17 is below the upper limit of the confidence​ interval, so we can conclude that the population proportion is less than 0.17.

Step-by-step explanation:

We are given that a survey of 270 young professionals found that one dash eighth of them use their cell phones primarily for​ e-mail.

Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;

                            P.Q. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of people who use their cell phones primarily for​ e-mail =  $\frac{1}{8}$ = 0.125

           n = sample of young professionals = 270

           p = population proportion

Here for constructing 95% confidence interval we have used One-sample z test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                   of significance are -1.96 & 1.96}  

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [$\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

        = [ $0.125-1.96 \times {\sqrt{\frac{0.125(1-0.125)}{270} } }$ , $0.125+1.96 \times {\sqrt{\frac{0.125(1-0.125)}{270} } }$ ]

        = [0.08 , 0.16]

Therefore, 95% confidence interval for the population proportion who use cell phones primarily for​ e-mail is [0.08 , 0.16].

Since, the above confidence interval have values which is less than 0.17; so we conclude that the population proportion is less than 0.17.