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, 278 278 babies were born to parents using the new? method, and 252 252 of them were boys. Use a 0.05 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. test statistic is z= round 2 decimal places

Explanation / Answer

Solution:

Here, we have to use Z test for population proportion.

The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: New method is not effective in increasing the likelihood that a baby will be a boy.

Alternative hypothesis: Ha: New Method is effective in increasing the likelihood that a baby will be a boy.

H0: p = 0.5 vs. Ha: p > 0.5

(We assume the natural proportion of baby boy and bay girl is same.)

This is an upper/right tailed (one tailed) test.

WE are given

Favourable obs. = X = 252

Sample size = N = 278

Sample Proportion = P = 0.90647482

Level of significance = ? = 0.05

Upper critical value = 1.6449 (by using z-table)

Test statistic formula is given as below:

Z = (P – p) / sqrt(p*(1 – p)/N)

Z = (0.90647482 – 0.5) / sqrt(0.5*(1 – 0.5)/278)

Z = (0.90647482 – 0.5) /0.0300

Z = 13.5546

Test statistic = Z = 13.55

P-value = 0.0000 (by using z-table or excel)

P-value < ? = 0.05

So, we reject the null hypothesis that new method is not effective in increasing the likelihood that a baby will be a boy.

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

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