Find attached A random sample of 2000 adult women indicated that 1041 of them asked for medical assistance the last time they felt sick. The same sized sample of men indicated that 993 of them asked for medical assistance.Use the z-values rounded to three decimal places to obtain the answers.a) Is it reasonable to conclude that women ask for medical assistance more often than men? Use α = 0.05. Find the P-value for this test.Round your answer to three decimal places (e.g. 98.765).P-value = [Input box]b) Suppose that \( p_1 = 0.55 \) and \( p_2 = 0.5 \). Determine the sample size needed to detect this difference with a probability of at least 0.95.n = [Input box]

The objective of the question is to test the hypothesis that women ask for medical assistance more often than men and to determine the sample size needed to detect a certain difference with a given probability.First, we need to calculate the sample proportions. For women, the sample proportion (p1) is 1041/2000 = 0.5205. For men, the sample proportion (p2) is 993/2000 = 0.4965.Next, we calculate the pooled sample proportion (p) which is (1041+993)/(2000+2000) = 0.5085.We then calculate the standard error (SE) using the formula SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] } where n1 and n2 are the sample sizes for women and men respectively. Substituting the values, we get SE = sqrt{ 0.5085 * (1 - 0.5085) * [ (1/2000) + (1/2000) ] } = 0.0112.Now, we calculate the z-score using the formula z = (p1 - p2) / SE. Substituting the values, we get z = (0.5205 - 0.4965) / 0.0112 = 2.143.The P-value is the probability that you have falsely rejected the null hypothesis. It can be found using the standard normal distribution table or a calculator. For a z-score of 2.143, the P-value is 0.016.Since the P-value is less than the significance level (α = 0.05), we reject the null hypothesis. Therefore, it is reasonable to conclude that women ask for medical assistance more often than men.For part b, we need to determine the sample size needed to detect a difference of p1 - p2 = 0.55 - 0.5 = 0.05 with a probability of at least 0.95. We use the formula n = [ (Zα/2 * sqrt[ 2p(1 - p) ] + Zβ * sqrt[ p1(1 - p1) + p2(1 - p2) ] ) / (p1 - p2) ]^2 where Zα/2 is the z-value that cuts off the upper α/2 area of the standard normal curve and Zβ is the z-value that cuts off the upper β area of the curve. For α = 0.05, Zα/2 = 1.96. For β = 1 - 0.95 = 0.05, Zβ = 1.645.Substituting the values into the formula, we get n = [ (1.96 * sqrt[ 2*0.525*(1 - 0.525) ] + 1.645 * sqrt[ 0.55*(1 - 0.55) + 0.5*(1 - 0.5) ] ) / 0.05 ]^2 = 3854. Therefore, a sample size of 3854 is needed to detect this difference with a probability of at least 0.95.Based on the calculations, it is reasonable to conclude that women ask for medical assistance more often than men as the P-value is less than the significance level. Also, a sample size of 3854 is needed to detect a difference of 0.05 with a probability of at least 0.95.