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

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

Explanation:

`H_0 : U_1 = U_2\\H_1 : U_1\\eq U_2`

`\alpha =0.05`

`t= \frac{m_1-m_2}{\sqrt{(S_1^(2) )/(n_1) +(s_2^(2) )/(n_2) } } =(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

Failed to Reject Null Hypothesis

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