Answer:
- The probability of a girl born to a couple using this technique = 0.935
- Yes, the technique does appear effective in improving the likelihood of having a girl baby.
Explanation:
1) In a test of gender selection, there are 200 girls and 14 baby boys.
To obtain the probability of a girl born to a couple using this technique
P(girls) = n(girls) ÷ n(total)
n(girls) = 200
n(total) = 200 + 14 = 214
P(girls) = (200/214) = 0.9346 = 0.935
2) Does it appear that the technique is effective in increasing the likelihood that a baby will be a girl?
We use an hypothesis test to confirm this. For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
Normally, the proportion of new girl babies and new boy babies should be close to each other (around 0.5 each), but this claim is that this gender selection technique favours the girl babies more than the male babies.
The null is that there is no significant evidence to conclude that the gender selection technique does favour more girl babies than boy babies.
The alternative hypothesis is that there is significant evidence to conclude that the gender selection technique does favour more girl babies than boy babies.
Mathematically,
The null hypothesis is represented as
H₀: p ≤ 0.50
The alternative hypothesis is given as
Hₐ: p > 0.50
To do this test, we will use the t-distribution because no information on the population standard deviation is known
So, we compute the t-test statistic
t = (x - μ)/σₓ
x = sample proportion = 0.935
μ = p₀ = The standard proportion we are comparing against = 0.50
σₓ = standard error = √[p(1-p)/n]
where n = Sample size = 214
p = 0.935
σₓ = √[0.935×0.065/214] = 0.0168521609 = 0.01685
t = (0.935 - 0.50) ÷ 0.01685
t = 25.81
checking the tables for the p-value of this t-statistic
Degree of freedom = df = n - 1 = 214 - 1 = 213
Significance level = 0.05 (most tests are performed at this level)
The hypothesis test uses a one-tailed condition because we're testing only in one direction.
p-value (for t = 25.81, at 0.05 significance level, df = 213, with a one tailed condition) = 0.000000001
The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.000000001
0.000000001 < 0.05
Hence,
p-value < significance level
This means that we reject the null hypothesis & say that there is enough evidence to conclude that the gender selection technique does favour more girl babies than boy babies.
So, yes, the technique does appear effective in improving the likelihood of having a girl baby.
Hope this Helps!!!