## Question 796:

1

No answer provided yet.For this questions we basically just need to lookup the t-statistic in a t-table to see if it is significant (p<.05). The only piece of information we need is the degrees of freedom (df). To find the df for this 2-sample t-test, we need to know whether we'd assume equal variances or unequal variances between the two populations. The difference in standard deviations is not very large and since it doesn't say either way, we'll use the simpler equal variance calculation for degrees of freedom : df = n1+n2 -2 = 338+119= 455.

We can use a t-table or the Percentiles from the t-Distribution Calculator and enter 1.99 and 455, we get the p-value of .0472, which is less than .05 so we conclude there is a difference between means based on the test statistic. If we were to assume unequal variances, the computation for the degrees of freedom would have gotten us 391 instead of 455. Even at this level the p-value is less than .048 so we still reject the Null Hypothesis and conclude there is a difference between very happy and not happy based on ideal number of children.

As a final note, the obtained t-statistic reported doesn't match the data. If you conduct a 2-sample t-test on the means and standard deviations given the large sample size, the t-statistic is 4.47 well above 1.99, and again also leading to the same conclusion--that there is a difference in the mean ideal number of children between very happy and not happy at all groups.