Question

A nutritionist is interested in the relationship between cholesterol and diet. The nutritionist developed a non-vegetarian and vegetarian diet to reduce cholesterol levels. The nutritionist then obtained a sample of clients for which half are told to eat the new non-vegetarian diet and the other half to eat the vegetarian diet for five months. The nutritionist hypothesizes that the non-vegetarian diet will increase cholesterol levels more. What can the nutritionist conclude with α = 0.05. Below are the cholesterol levels of all the participants after five months.

non- vegetarian	vegetarian
106 121 141 146 156 196 106 106	126 171 196 108 231 256 131 196

If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[   ,   ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and/or select "na" below.
d =   ;  ---Select--- na trivial effect small effect medium effect large effect
r² =   ;  ---Select--- na trivial effect small effect medium effect large effect

Answer 1

From the given data

	Non- Vegetarian(X1)	Vegetarian(X2)
	106	126
	121	171
	141	196
	146	108
	156	231
	196	256
	106	131
	106	196
Mean(X1)	134.75
Mean(X2)	176.875
s1	31.70737
s2	52.63469

and the sample sizes are n1​=8 and n2​=8.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ < μ2​

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df = df=14. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this left-tailed test is tc​=−1.761, for α=0.05 and df=14.

The rejection region for this left-tailed test is R={t:t<−1.761}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t=−1.939<tc​=−1.761, it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1​ is less than μ2​, at the 0.05 significance level. which means that the non-vegetarian diet will increase cholesterol levels more.

Confidence Interval

t* is critical value = 1.761

The 95% confidence interval is −88.72<μ1​−μ2​<4.47.

e) Compute the corresponding effect size(s) and indicate magnitude(s).

Cohen's d = (M₂ - M₁) ⁄ SD_pooled

SD_pooled = √((SD₁² + SD₂²) ⁄ 2)

Cohen's d = (176.875 - 134.75) ⁄ 43.449234 = 0.969522.