Question

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were​ born, and 340 of them were girls. Use the sample data to construct a 99​% confidence interval estimate of the percentage of girls born. Based on the​ result, does the method appear to be​ effective? nothingless than pless than nothing ​(Round to three decimal places as​ needed.) Does the method appear to be​ effective? Yes​, the proportion of girls is significantly different from 0.5. No​, the proportion of girls is not significantly different from 0.5.

Answer 1

Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes​, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were​ born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% 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 girls born = $\frac{340}{400}$ = 0.85

             n = sample of babies = 400

             p = population percentage of girls born

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

                                                    of significance are -2.58 & 2.58}  

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [$\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

= [ $0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ , $0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }$ ]

 = [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Let p = population proportion of girls born.

So, Null Hypothesis, $H_0$ : p = 0.50      {means that the proportion of girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p $\neq$ 0.50      {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is One-sample z proportion test statistics;

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

where, $\hat p$ = sample proportion of girls born = $\frac{340}{400}$ = 0.85

             n = sample of babies = 400

So, the test statistics  =  $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

                                     =  19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.