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 boy. As of this writing, 294 babies were born to parents using the new method, and 237 of them were boys. 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 boy. 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

Explanation / Answer

Null and Alternative Hypothesis:

H0: P 0.5

H1: P > 0.5

Test Statistics:

z = (p^ -P )/sqrt ((P*(1-P))/n)

p^ = 237 / 294 = 0.8061

n = 294

z = (0.8061-0.5)/sqrt ((0.5*(1-0.5))/294) [ sqrt = square root ]

= 10.50

p-value = P ( z > 10.50 )

= 1 – P ( z < 10.5)

= 1 – 1                            [Using Normal Distribution Table]

= .0000

Level of significance (alpha) = .05

Here the p-value is less than the level of significance (.0000 < .05); the null hypothesis can be rejected at 5% level of significance.

There is sufficient evidence to conclude that the new method is effective in increasing the likelihood that a baby will be a boy.

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