posted by Mike .
Data from the latest census was analyzed. It was determined that one-parent households spend less on food away from home that do two-parent households. The mean amount spent by 30 one-parent households is $1876 and the standard deviation is $113. The mean amount spent by 30 two-parent households is $1878 and the standard deviation is $85. At a a=0.05 can you support this conclusion?
Try an independent groups t-test.
Ho: µ1 = µ2 -->null hypothesis
Ha: µ1 < µ2 -->alternate hypothesis
Use (n1 + n2 - 2) degrees of freedom for this test. Use a t-table to determine your cutoff or critical value to reject the null using 0.05 level of significance for a one-tailed test. If your test statistic exceeds the critical value from the table, reject the null and conclude a difference (µ1 < µ2). If the test statistic does not exceed the critical value from the table, do not reject the null.
I hope this will help get you started.