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 conceiving a girl. As of this writing, 954 babies were born to parents using the new method, and 884 of them were girls. Use a 0.05 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 following is the hypothesis test to be conducted? What is the test statistic? Z = What is the P=value? What is the conclusion about the null hypothesis? What is the conclusion?

Explanation / Answer

Total babies = 954

GIrls = 884

Null hypothesis : H0 : Probability of conceiving a baby girl is same with the new method as earltier. p = 0.5

ALternative Hypotheis : Ha : Probability of conceiving a baby girl is higher with the new method as earltier. p > 0.5

Here sample proportion p^ = 884/954 = 0.9266

Here standard error of the proportion se0 = sqrt[p0(1-p0)/N] = sqrt [0.5 * 0.5/ 954] = 0.0162

Test statistic

Z = (p^ - p0 )/ se0 = (0.9266 - 0.5)/ 0.0162 = 26.33

P -value = Pr(Z > 26.33) = 0.000 < 0.05

as we shall reject the null hypothesis and can conclude that there is significant increase in likelihood of a girl getting born with the new method.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.