Question

If the weight of male and females were the same then the difference would be zero. Based on the confidence interval, are males and females the same average weight? Why or why not? Explain.

Observed Samples Biological SexN Mean StDev Variance Minimum Median Maximum Female Male 286 64.6556 2.9682 8.8103 48.0000 64.7500 72.0000 234 70.4282 3.1015 9.6193 59.0000 71.0000 79.0000 Bootstrap Histogram for Weight (in lbs) by Biological Sex 95% Confidence Interval 6.28858 5.77140 -5.28077 90 80 70 60 S 50 2 40 30 20 10 -6.50 -6.25 15,000 585 5.50 5.25 .ti.@i) Difference in Means

Answer 1

Let x₁,x₂.....x_n1 and x₁,x₂.....x_n2 are random sample of sizes n₁ and n₂ respectively. Let $\bar{X}$₁ and $\bar{X}$2  denotes the sample means of first and second samples

Let $\sigma_{_{_{_{_{1}}}}}$ and $\sigma_{_{_{_{_{2}}}}}$ denotes the population standard deviations and S₁ and S₂ denotes the sample standard deviations

In the given problem we have to test the null hypothesis that the two population means are the same against the alternative that they are not same

i.e H₀:$\mu _{1}=\mu_{2}$ against $H_{1}:\mu _1\neq \mu_2$ where $\mu _{1}and \ \mu_{2}$ are the population means.

Under the null hypothesis the test statistic is,

$\frac{{\bar{X}{_{1}}-\bar{X}_{2}}}{\sqrt{\left (S _{1}^2/n_{1} +S _{2}^2/n_{2} \right )}}$

=$\frac{64.6556-70.4282}{\sqrt{{\frac{8.8103}{286}+\frac{9.6193}{234}}}}$

=-21.5281

which can be denoted by Z and Z follows a Standard Normal distribution

The critical region corresponding to the alternative hypothesis is $\begin{vmatrix}Z \end{vmatrix}\geq Z_{\alpha /2}$

Hence $\begin{vmatrix}Z \end{vmatrix}=21.5281\geq 1.96=Z_{\alpha /2}$ where$Z_{\alpha /2}$ can be determined from standard normal tables for $\alpha =0.05$

Since the value of the test statistic lies in the critical region we reject the null hypothesis and conclude that the average weight of male and female is not equal