Question

In a study comparing calories in vegetarian and non vegetarian entrees the average number of calories in 28 vegetarian entrees was found to be Y¯ = 351 with a standard deviation S = 119 while the number of calories in 28 non vegetarian entrees was Y¯ = 305 with S = 125.
A) Find a 99% confidence interval for µV − µNV.
B) Test at α = .05 that vegetarian entrees have more calories. State the hypothesis, test statistic and your conclusion.
C) If non vegetarian meals have on average 25 more calories (which all other things being equal would lead to a 7.8 kg weight increase per year), how large do the samples need to be to detect such an increase with power .8 at the above level of significance α = .05. Use the statistics obtained above as the parameter values needed for the power calculation.

Answer 1

Solution:

Calculating the polled variance

$S_p^2 = \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}$

    = $\frac{(28-1)\times 119^2+(28-1)\times 125^2}{28+28-2}$

   = 14893

Now we calculate the estimated standard error

$S_{M1-M2} =\sqrt {\frac{S_p^2}{n_1}+\frac{S_p^2}{n_2}}$

            $=\sqrt{\frac{14893}{28}+\frac{14893}{28}} = 32.61$

Thus , the 99 percent confidence interval is

= $(M_1-M_2) \pm t_{54.001}\times S_{M_1-M_2}$

= 351 - 305 ± 2.660 x 32.6

= (46) ± 86.74

= (-40.74, 132.74)

Now the null hypothesis is :

$H_0 : u_1-u_2 = 0$

Against the alternative hypothesis

$H_0 : u_1-u_2 > 0$

Computing the statistics ,

$t = \frac{(M_1-M_2)-(u_1-u_2)}{S_{M_1-M_2}}$

 $= \frac{(351-305)-0}{32.61}$

 = 1.41