Question

A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first stages of a clinical trial, it was successful for 7 out of the 14 women. What is the 95% confidence interval for p, the true proportion of all women who will find success with this new treatment?

Answer 1

Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true 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 women who find success with this new treatment = $\frac{7}{14}$ = 0.50

          n = sample of women = 14

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

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

= [ $0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ , $0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } }$ ]

 = [0.238 , 0.762]

Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].