Question

Nutritional researchers conducted an investigation of two high-fiber diets intended to reduce cholesterol level. a group of people with high cholesterol were randomly selected to receive an "rice" diet or a "vegetable" diet for 20 days. Use t test to compare the diets at 5% significance level. Null Hypothesis: The means from two respective diets are the same Alternative Hypothesis: The two means from the two diets differ.

Answer 1

Answer:

Failed to reject $H_0$, hence the two means from the two diet differs.

Step-by-step explanation:

$H_0 : U_1 = U_2\\H_1 : U_1\neq U_2$

$\alpha =0.05$

$t= \frac{m_1-m_2}{\sqrt{\frac{S_1^{2} }{n_1} +\frac{s_2^{2} }{n_2} } } =\frac{2.21}{1.35} = 1.637$

t has v degrees of freedom and

$v= 36$

$t\alpha , v = tx_{0.05} , 36 = 1.645$

Decision Rule: Reject $H_0$ if $1.637 > 1.645$, otherwise do not reject.

Conclusion: Since the calculated test statistic is less than the critical point, we fail to reject the null hypothesis and therefore conclude that the average effects of the two high-fiber diets differ significantly at 5% level of significance.

Answer 2

Answer:

Failed to Reject Null Hypothesis

Step-by-step explanation:

Null Hypothesis: The means from two respective diets are the same Alternative Hypothesis:  The two means from the two diets differ.

Fall in Cholesterol (mg/dL)

Diet                     n     Mean      SD

Rice                    7       14.4       7.8

Vegetable          9       12.19     7.86