Please help!!!2 **Smoking Survey**National statistics show that 23% of men smoke and 18.5% of women do. A random sample of 145 men indicated that 33 were smokers, and of 114 women surveyed, 12 indicated that they smoked.**Part: 0 / 2***Part 1 of 2*Construct a 96% confidence interval for the true difference in proportions of male and female smokers. Use \( \hat{p}_1 \) for the proportion of men who smoke. Round your answers to three decimal places.\[ 22 < \hat{p}_1 - \hat{p}_2 < 22 \]**Part: 1 / 2***Part 2 of 2*Comment on your interval—does it support the claim that there is a difference?(Choose one) since 0 (Choose one) contained in the interval, it allows for the possibility that there is (Choose one) between the proportions of men and women who smoke.There are no graphs or diagrams in the image. The focus is on constructing a confidence interval and interpreting its significance.

Solution: Given information: n1= 145x1=33p^1=x1n1=33145=0.23n2=114x2=12p^2=x2n2=12114=0.105The 96% confidence interval for the difference in proportions between males and female is (p^1-p^2)±Zα/2×p^1×(1-p^1)n1+p^2×(1-p^2)n2C=0.96 confidence level α=1-C=0.04Zα/2=Z0.02=2.0537 From Excel (0.23-0.105)±2.0537×0.23×0.77145+0.105×0.8951140.125±2.0537×0.001221+0.000824(0.125±2.0537×0.002045)(0.125±2.0537×0.045225)(0.125±0.092880)(0.032120, 0.217880)(0.032, 0.218) The 96% confidence interval for the difference in proportions between males and female is 0.032<p1-p2<0.218 part 2) There is sufficient evidence to support the claim that there is a difference since 0 is not contained in the interval it allows for the possibility that there is a difference between the proportions of men and women who smoke.