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, 279 babies were born to parents using the new method, and 236 of them were boys. 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 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

As, 279 babies were born to parents using the new method, and 236 of them were boys.

p = 236 / 279 = 0.8459 is the proportion of baby boys in total number of babies.

Here we want to test that "The new method is effective in increasing the likelihood that a baby will be a boy"

So null hypothesis is H0 : The new method is not effective in increasing the likelihood that a baby will be a boy.

and alternative hypothesis is H1 : The new method is effective in increasing the likelihood that a baby will be a boy

We can used one sample proportion z test.

Let' write given information.

n = sample size = 279

x = number of favourable events = 236

level of significance = 0.01

Using minitab.

The command for one sample proportion z test in minitab is

Stat>>>Basic statistics>>>1-proportion...

Then click on summarized data

number of events = x = 236

Number of trials = n = 279

Click on "Perform hypothesis test

Hypothesized proportion P = 0.5   (because under the null hypothesis we claim that new method is not effective so we test proportion of baby boy and baby girl will same)

then click on option

Level of confidence in percentage = c = ( 1- lpha)*100 = = (1 -0.01)*100 = 99.0

so put "Confidence level " = 99.0

Alternative = not equal

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

MTB > POne 279 236;
SUBC>   Test 0.5;
SUBC>   Confidence 99;
SUBC>   UseZ.

Test and CI for One Proportion

Test of p = 0.5 vs p not = 0.5

Sample    X     N    Sample p         99% CI        Z-Value P-Value
1      236 279 0.845878 (0.790198, 0.901558)    11.55    0.000

Using the normal approximation.

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.00 < 0.01 so we used 1st rule.

That is we reject null hypothesis

Conclusion: At 1% level of significance there are sufficient evidence to say that the new method is effective in increasing the likelihood that a baby will be a boy.

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