Answer to Question #186038 in Statistics and Probability for Olivia Padula

Question

 A social psychologist is interested in gender differences in the sociability of teenagers. Using the number of good friends as an indicator of sociability, the researcher found that females had on average 5.9 close friends (s=3.1; n=44) and males had on average 4.8 close friends (s=2.6; n=35). Is the difference in the sociability between males and

females statistically significant? Set alpha to 0.02 find Null and research hypothesis 1 or two tailed? Find Critical value? find test statistic?

Answer 1

$To \ test \ gender \ differences \ in \ the \ sociability \ of \ teenagers.\\$ $Using \ the \ number \ of \ good \ friends \ as \ an \ indicator \ of \ sociability.\\$

$Females \ had \ on \ average \ 5.9 \ close \ f riends \ (s=3.1; n=44)$

$Males \ had \ on \ average \ 4.8 \ close \ friends \ (s=2.6; n=35)\\ \Rightarrow Let \ \bar x_1 =5.9, \ s_1 =3.1 \ \& \ n_1 = 44\\ \bar x_2 =4.8, \ s_2 =2.6 \ \& \ n_2 = 35\\ Let \ the \ null \ hypothesis \ be \ H_0 \ : \ There \ is \ no \ significant \ difference \\ of \ sociabilty \ in \ male \ \& \ female\ \Rightarrow \mu_1 =\mu_2 \\ The \ alternative \ hypothesis \ be\ H_1: \mu_1 \ne \ \mu_2 \\ The \ test \ statistic \ for \ the \ difference \ of \ means \ is \\ z=\frac{\bar x_1-\bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{5.9-4.8}{\sqrt{\frac{(3.1)^2}{44}+\frac{(2.6)^2}{35}}}\\ =\frac{1.1}{\sqrt{0.2184090909+0.1931428571}}\\ =\frac{1.1}{\sqrt{0.411551948}}\\ =0.64152\\ \Rightarrow z=0.64 (approximately)\\ For \ the \ significance \ leve \ \alpha = 0.02 \\ the \ critical \ value \ is \ z_{\frac{\alpha}{2}}=z_{0.01}=2.33 \\ (for \ two \ tailed \ test )\\ The \ test\ statistic \ z \lt \ 2.33\\ Therefore, there \ is \ no \ reason \ to \ reject \ the \ null \ hypothesis.\\ Hence, we \ can \ conclude \ that \ there \ is \ no\ significant \\ difference \ in \ sociability \ of \ male \ and \ female\ at \ 0.02 \\ significance \ level, \ by \ two \ tailed \ test.$