Researchers randomly divided 500 volunteers into two equal-sized groups. one group spent 6 months on the paleo diet. a control group received a pamphlet about controlling portion sizes. during the study, the paleo group lost on average 7 pounds (standard deviation of 20 pounds) while the control group had lost on average 5 pounds (standard deviation of 12 pounds). calculate the p-value assessing if the paleo diet provides more weight control than the control treatment.

And we can use the following excel code to find it: "=1-T.DIST(1.356,498,TRUE)"

Step-by-step explanation:

Data given and notation

represent the mean for the sample Paleo

represent the mean for the sample Control

represent the sample standard deviation for the sample Paleo

represent the sample standard deviation for the sample Control

sample size selected Paleo

sample size selected Control

represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the Paleo diet provides more weight control than the control treatment, the system of hypothesis would be:

Null hypothesis:

Alternative hypothesis:

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

(1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

P-value

The first step is calculate the degrees of freedom, on this case:

Since is a one side right test the p value would be:

And we can use the following excel code to find it: "=1-T.DIST(1.356,498,TRUE)"

step-by-step explanation:

4227473658779875563886538705958641728563470543621287463824132573847969p05749365273847906970680574936214

step-by-step explanation:

mean or average is 40.6