A dietician wants to know if the cost per month of the Wonder Diet is more expensive than the Southwest Diet. Healthy Eating Magazine published the population standard deviations as $89 for the Wonder Diet and $75 for the Southwest Diet. She conducted a random sample of 20 clients on each diet. The mean amount for the Wonder Diet was $643 and the Southwest Diet was $588. At a = 0.01, can the dietician say that the Wonder Diet is more expensive than the Southwest Diet? Assume the populations are normally distributed.

We are given that Healthy Eating Magazine published the population standard deviations as $89 for the Wonder Diet and $75 for the Southwest Diet. She conducted a random sample of 20 clients on each diet. The mean amount for the Wonder Diet was $643 and the Southwest Diet was $588.

We have to test whether the Wonder Diet is more expensive than the Southwest Diet or not.

Firstly, let the Population mean amount of wonder diet be [tex]\mu_1[/tex]

and the Population mean amount of Southwest diet be [tex]\mu_2[/tex] .

So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 \leq \mu_2[/tex] or [tex]\mu_1 -\mu_2 \leq 0[/tex] {means that the Wonder Diet is less expensive than or equal to the Southwest Diet}

Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu_1 > \mu_2[/tex] or [tex]\mu_1 - \mu_2>0[/tex] {means that the Wonder Diet is more expensive than the Southwest Diet}

The test statistics that will be used here is Two-sample z-test statistics,i.e;

T.S. = [tex]\frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^{2}_1 }{n_1}+\frac{\sigma^{2}_2}{n_2} } }[/tex] ~ N(0,1)

where, [tex]\bar X_1[/tex] = sample mean amount for the Wonder Diet = $643

[tex]\bar X_2[/tex] = sample mean amount for the Southwest Diet = $588

[tex]\sigma_1[/tex] = population standard deviation for Wonder diet = $89

[tex]\sigma_2[/tex] = population standard deviation for Southwest diet = $75

[tex]n_1[/tex] = sample of clients for Wonder diet = 20

[tex]n_2[/tex] = sample of clients for Southwest diet = 20

So, test statistics = [tex]\frac{(643-588)-(0)}{\sqrt{\frac{89^{2}}{20}+\frac{75^{2}}{20} } }[/tex]

= 2.113

Now, at 0.01 significance level z table gives critical value of 2.3263. Since our test statistics is less than the critical value of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that at a = 0.01, the dietician can't say that the Wonder Diet is more expensive than the Southwest Diet.