A study shows that the probability of a baby being born with gene x is 0.6. The probability of a baby being born with gene y is 0.7.
(a)What is the probability of being born with both genes assuming that the probabilities are independent?
(b)In fact; 90% of those born with gene x also have gene y. Of those born with gene y; what proportion also have gene x?
(c)Comment on the relative sizes of the confidence intervals for P(x) and P(y).
(d)If p(x) comes from a sample of 140 and p(y) from a sample of 210 are the true population proportions statistically different? State your hypotheses clearly and show all workings.
(e)Assuming p(x) is known to be true, what sample size would be required to make P(y) distinguishable from P(x) at 95% confidence? Show your workings.
(a) If the probabilities of gene X and gene Y are independent, then the probability of being born with both genes is simply the product of their individual probabilities:
P(X and Y) = P(X) * P(Y)
= 0.6 * 0.7
= 0.42
So the probability of being born with both genes assuming independence is 0.42.
(b) Given that 90% of those born with gene X also have gene Y, we can determine the proportion of those born with gene Y who also have gene X using conditional probability.
P(X|Y) = P(X and Y) / P(Y)
Since we know P(X and Y) from part (a) as 0.42, and the probability of being born with gene Y is given as 0.7, we can calculate:
P(X|Y) = 0.42 / 0.7
= 0.6
So, of those born with gene Y, the proportion that also have gene X is 0.6.
(c) The confidence intervals for P(X) and P(Y) will depend on the sample sizes and the level of confidence chosen. Without specific sample information, it is difficult to compare the relative sizes of the confidence intervals.
(d) To determine if the true population proportions, p(X) and p(Y), are statistically different, we can conduct a hypothesis test. The null hypothesis (H0) assumes that the true population proportions are equal, while the alternative hypothesis (Ha) assumes they are not equal.
H0: p(X) = p(Y)
Ha: p(X) ≠ p(Y)
We can use the two-sample proportion test to compare the proportions. The test statistic follows an approximately standard normal distribution under the null hypothesis.
The test statistic can be calculated as:
z = (p̂(X) - p̂(Y)) / sqrt((p̂(X)*(1-p̂(X))/n(X)) + (p̂(Y)*(1-p̂(Y))/n(Y)))
Where p̂(X) and p̂(Y) are the sample proportions, n(X) and n(Y) are the respective sample sizes.
We can then calculate the p-value associated with the test statistic using the standard normal distribution. If the p-value is less than the chosen significance level (e.g., 0.05), we reject the null hypothesis and conclude that the true population proportions are statistically different.
(e) To determine the sample size required to make P(Y) distinguishable from P(X) at a 95% confidence level, we need to perform a power analysis. A power analysis takes into account the desired level of confidence, the effect size, and the desired power level.
Since p(X) is known to be true, we can use it as the baseline for comparison. We need to determine the sample size that will result in a statistically significant difference from p(X) when comparing to p(Y).
Step 1: Determine the effect size (d):
d = |p(Y) - p(X)| / sqrt((p(X)*(1-p(X)) + (p(Y)*(1-p(Y))))
= |p(Y) - p(X)| / sqrt(p(X)*(1-p(X)))
Step 2: Use a sample size calculator or power analysis formula to find the required sample size. The formula depends on the specific statistical test and software being used.
For example, using a two-sample proportion test, the sample size formula can be calculated as:
n = (Zα/2 + Zβ)^2 * (p(X)*(1-p(X)) + d^2) / d^2
Where Zα/2 and Zβ are the corresponding Z-scores for the desired level of significance (e.g., 95% confidence) and power (e.g., 80% power), respectively.
By plugging in the values, the required sample size can be calculated.
(a) To find the probability of being born with both genes assuming they are independent, you can simply multiply the individual probabilities together.
Probability of being born with both genes (P(x and y)) = P(x) * P(y) = 0.6 * 0.7 = 0.42
Therefore, the probability of being born with both genes assuming independence is 0.42.
(b) If 90% of those born with gene x also have gene y, it means that the probability of having y given x (P(y|x)) is 0.9.
To find the proportion of those born with gene y who also have gene x (P(x|y)), you can use Bayes' theorem:
P(x|y) = (P(y|x) * P(x)) / P(y)
Given that P(y|x) is 0.9 and P(x) is 0.6, we need to find P(y).
P(y) = 1 - P(not y) = 1 - (1 - P(x)) = P(x) = 0.6
Now we can calculate P(x|y):
P(x|y) = (0.9 * 0.6) / 0.6 = 0.9
Therefore, 90% of those born with gene y also have gene x.
(c) The size of the confidence intervals for P(x) and P(y) depends on the sample sizes and the variability within the samples. Without more information, it is not possible to comment on the relative sizes of the confidence intervals for P(x) and P(y).
(d) To test if p(x) from a sample of 140 and p(y) from a sample of 210 are statistically different, we can use a two-proportion z-test.
The null hypothesis (H0) is that there is no difference between the two population proportions:
H0: p(x) = p(y)
Alternative hypothesis (Ha) is that there is a difference between the two population proportions:
Ha: p(x) ≠ p(y)
To test this, we calculate the test statistic (z-value):
z = (p̂(x) - p̂(y)) / sqrt((p(x)*(1-p(x))/n(x)) + (p(y)*(1-p(y))/n(y)))
where p̂(x) and p̂(y) are the sample proportions, n(x) and n(y) are the sample sizes, and p(x) and p(y) are the estimated population proportions.
We compare the test statistic to the critical value at the desired level of significance (e.g., 0.05) to determine if we can reject the null hypothesis.
If the test statistic falls outside the critical value range, we reject the null hypothesis and conclude that the population proportions are statistically different. If the test statistic falls within the critical value range, we fail to reject the null hypothesis, meaning that there is not enough evidence to support a significant difference between the proportions.
(e) To determine the sample size required to make P(y) distinguishable from P(x) at 95% confidence, you can use the formula for sample size calculation for proportions:
n = (z^2 * p̂(1 - p̂)) / E^2
where n is the required sample size, z is the z-value at the desired level of confidence (1.96 for 95% confidence), p̂ is the estimated population proportion (0.6 for P(x)), and E is the desired margin of error.
Assuming you want a margin of error (E) of 0.05, the calculation becomes:
n = (1.96^2 * 0.6(1-0.6)) / (0.05^2)
n = (3.8416 * 0.24) / 0.0025
n = 365.664
Therefore, the sample size required to make P(y) distinguishable from P(x) at 95% confidence is approximately 366.