5- Construct a 90% confidence interval estimate for the mean * using the sample information n =21, * =13.6, and s =2.4. Step 1 is solved: t (df, "/ 2 ) = (20, 0.05) = 1.72 FORMULA FOR CONFIDENCE INTERVAL ESTIMATE USING T DISTRIBUTION E = t (df, 0/2 ) (s/ Jm )= GIVEN THE ABOVE INFORMATION: CALCULATE THE STANDARD ERROR & CONSTRUCT THE 90% CONFIDNECE INTERVAL ESTIMATE (PTS: 3) ANS: 6- A researcher is interested in testing to determine if the mean price of a casual lunch is different in the city than it is in the suburbs. The null hypothesis is that there is no difference in the population means (i.e. the difference is zero). The alternative hypothesis is that there is a difference (i.e. the difference is not equal to zero). He randomly selects a sample of 9 lunch tickets from the city population resulting in a mean of $14.30 and a standard deviation of $3.40. He randomly selects a sample of 14 lunch tickets from the suburban population resulting in a mean of $11.80 and a standard deviation $2.90. He is using an alpha value of .10 to conduct this test. Assuming that the populations are normally distributed and that the population variances are approximately equal, the degrees of freedom for this problem are SOLVE AND SHOW THE STEPS (PTS: 3) ANS