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 th

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.

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Anna 1 month 2021-10-22T07:40:09+00:00 1 Answer 0 views 0

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    2021-10-22T07:41:48+00:00

    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.

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