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More Information From the Online Dating Survey A survey conducted in July 2015 asked a random sample of American adults whether

svet-max [94.6K]

Answer:

90% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating is [0.095 , 0.148].

Step-by-step explanation:

We are given that a survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating.

The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating.

Firstly, the pivotal quantity for 90% 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 adults who said that they had used online dating = $\frac{50}{411}$ = 0.122

n = sample of adults between the ages of 55 and 64 = 411

p = population proportion of all US adults ages 55 to 64 to use online dating

Here for constructing 90% confidence interval we have used One-sample z proportion statistics.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.122-1.645 \times {\sqrt{\frac{0.122(1-0.122)}{411} } }$ , $0.122+1.645 \times {\sqrt{\frac{0.122(1-0.122)}{411} } }$ ]

= [0.095 , 0.148]

Therefore, 90% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating is [0.095 , 0.148].

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