A clinical trial tests a method designed to increase the probability of conceiving a girl. In a study 306 babies were born, and 174 of them were girls. the sample data with a 0.01 significance level to test the claim that with this method, the
probability of being a girl is greater than 0.5.
You can try a proportional one-sample z-test for this one since this problem is using proportions.
Here's a few hints to get you started:
Null hypothesis:
Ho: p = .5 -->meaning: population proportion is equal to .5
Alternative hypothesis:
Ha: p > .5 -->meaning: population proportion is greater than .5 (this is a one-tailed test)
Using a formula for a proportional one-sample z-test with your data included, we have:
z = (.57 - .5) -->test value (174/306 is approximately .57) minus population value (.5) divided by
√[(.5)(.5)/306]
Finish the calculation. Check a z-table for .01 level of significance for a one-tailed test. Remember if the null is not rejected, then there is no difference. If the null is rejected, then p > .5 and there is a difference.
I hope this will help get you started.
To test the claim that the probability of having a girl is greater than 0.5 using the given sample data, we can perform a hypothesis test. This will help us determine if there is enough evidence to support the claim or not.
Here are the steps to perform the hypothesis test:
Step 1: State the null hypothesis (H0) and alternative hypothesis (Ha):
The null hypothesis states that the probability of having a girl is equal to or less than 0.5. In this case, H0: p ≤ 0.5.
The alternative hypothesis states that the probability of having a girl is greater than 0.5. In this case, Ha: p > 0.5.
Step 2: Set the significance level (α):
The significance level, denoted by α, determines the threshold for accepting or rejecting the null hypothesis. In this case, the significance level is given as 0.01 (or 1%).
Step 3: Calculate the test statistic:
For this hypothesis test, we will use a one-sample proportion z-test. The test statistic formula is:
z = (p̂ - p0) / √(p0(1-p0)/n),
where p̂ is the sample proportion, p0 is the hypothesized population proportion, and n is the sample size.
In this case, p̂ (sample proportion) is calculated as the number of girls (174) divided by the total number of babies (306):
p̂ = 174/306 ≈ 0.5686.
Substituting the given values in the formula, we get:
z = (0.5686 - 0.5) / √(0.5(1-0.5)/306).
Step 4: Determine the critical value:
To determine the critical value, we need to find the z-value corresponding to the specified significance level. Since we are testing if the probability of having a girl is greater than 0.5, this is a one-tailed test.
Using a z-table or a statistical software, we find that the critical value for a one-tailed test at the 0.01 significance level is approximately 2.33.
Step 5: Make a decision:
If the calculated test statistic (z) is greater than the critical value (2.33), we can reject the null hypothesis. If it is less than the critical value, we fail to reject the null hypothesis.
Step 6: Calculate the p-value:
The p-value is the probability of obtaining a test statistic as extreme as the one observed, assuming the null hypothesis is true. In this case, we calculate the p-value using the cumulative distribution function (CDF) of the standard normal distribution.
For a one-tailed test (greater than), the p-value is calculated as 1 minus the CDF value at the test statistic.
Step 7: Interpret the results:
If the p-value is less than the significance level (α), we reject the null hypothesis. If the p-value is greater than the significance level, we fail to reject the null hypothesis.
Now, you have all the information required to perform the hypothesis test. Plug in the values and compare the test statistic (z) with the critical value to make a decision regarding the null hypothesis.