To determine if the proportion of men with the blood disorder is significantly greater than the proportion of women with the disorder, we can perform a two-sample proportion z-test.
First, we need to calculate the sample proportions for both males and females:
For males:
Sample proportion (p1) = 41/140 = 0.293
For females:
Sample proportion (p2) = 30/110 = 0.273
Next, we determine the test statistic using the formula:
z = (p1 - p2) / √(p̂(1-p̂) * (1/n1 + 1/n2))
Where:
p̂ = (n1 * p1 + n2 * p2) / (n1 + n2)
n1 and n2 are the sample sizes for males and females, respectively.
For this case:
n1 = 140, n2 = 110
p̂ = (140 * 0.293 + 110 * 0.273) / (140 + 110) = 0.280
Now, we can calculate the test statistic:
z = (0.293 - 0.273) / √(0.280 * (1-0.280) * (1/140 + 1/110))
z ≈ 0.020 / √(0.280 * 0.720 * (0.0071 + 0.0091))
z ≈ 0.020 / √(0.280 * 0.720 * 0.0162)
z ≈ 0.020 / √(0.00873728)
z ≈ 0.020 / 0.0934525
z ≈ 0.214
The test statistic is approximately 0.214.
To determine if this test statistic is significant at a 0.01 level of significance, we compare it to the critical z-value for a 0.01 (α = 0.01) level of significance. We can use a standard normal distribution table or a calculator to find that the critical z-value for a two-tailed test at a 0.01 level of significance is approximately ±2.576.
Since 0.214 is within the range of -2.576 to +2.576, we fail to reject the null hypothesis. Thus, we cannot conclude at a 0.01 level of significance that the proportion of men with the blood disorder is significantly greater than the proportion of women with the disorder.
A random sample of 140 males shows that 41 males have a certain minor blood disorder, while a random sample of 110 females shows that only 30 have the disorder. Can we conclude at a 0,01 level of significance that the proportion of men in the population with this blood disorder is significantly greater than the proportion of women with this disorder. Give the test statistic.
1 answer
1 answer