A clinical trial was conducted using a new method designed to increase the proba

Question

A clinical trial was conducted using a new method designed to increase the probability of concelving a girl As of this writing, 926 bables were born to parents using the new method, and 862 of them were girls. Use a 0.01 significance level to test the claim that the new method is effective in increasing the likelihood that a baby will be a girl. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution. Which of the folowing is the hypothesis test to be conducted? H p 05 D. Ho p=0.5 H p>0.5 OF Ho P05 H P-05 H P 05 H1 , p#0.5 Hop#05 H p 05 What is the test statistic? OE. (Round to two decimal places as needed) What is the P-value? P-value (Round to four decimal places as needed.)

Explanation / Answer

1) option D) is correct

Ho p = 0.5

Ha : p > 0.5

as want to know if proportion of girl has increased

2) X = 862 , n = 926

p^ = X/n = 862/926 = 0.930885

Z = (p^ - p)/sqrt(pq/n) = (0.930885-0.5)/sqrt(0.5*0.5/926)

= 26.223874

p-value = P(Z> 26.223874) = 0.00000

since p-value < 0.05 (level of signifiace)   

we reject the null

option C) is correct

and conclude that there is enough evidence that the propotion of girl has increased

option A) is correct

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