Question

An experiment was planned to compare the mean time (in days) required to recover from a common cold for persons given a daily dose of 4 milligrams (mg) of vitamin C, μ2, versus those who were not, μ1. Suppose that 33 adults were randomly selected for each treatment category and that the mean recovery times and standard deviations for the two groups were as follows.

no vitamin suppl. 4 mg vitamin C
sample size 35 35
sample mean 6.9 5.8
sample standard deviation 2.9 1.2
A) suppose your research objective is to show that the use of vitamin C reduces the mean time required to recover from a common cold and its complications. Give the null and alternative hypothesis for the test. is this a one to one or a two tailed test?
B) Conduct the statistical test of the null hypothesis in part a and state your conclusion. Test using alpha= .05

Answer 1

`t=\frac{5.8-6.9}{\sqrt{(1.2^2)/(35)+(2.9^2)/(35)}}}=-2.074`

`df=n_(1)+n_(2)-2=35+35-2=68`

`p_v =P(t_((68))<-2.074)=0.0209`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we have enough evidence to reject the null hypothesis and then we can conclude that the mean for the group with the vitamin C is lower than the mean for the group without vitamin C

Explanation:

Data given and notation

`\bar X_(1)=6.9` represent the mean for the sample for no vitamin supply

`\bar X_(2)=5.8` represent the mean for the sample for 4 mg witamin

`s_(1)=2.9` represent the sample standard deviation for no vitamin

`s_(2)=1.2` represent the sample standard deviation for 4 mg vitamin

`n_(1)=35` sample size selected for 1

`n_(2)=35` sample size selected for 2

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the use of vitamin C reduces the mean time required to recover from a common cold and its complications, the system of hypothesis would be:

Null hypothesis:
`\mu_(2) \geq \mu_(1)`

Alternative hypothesis:
`\mu_(2) < \mu_(1)`

If we analyze the size for the samples both are > 30 but 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:

`t=\frac{\bar X_(2)-\bar X_(1)}{\sqrt{(s^2_(2))/(n_(2))+(s^2_(1))/(n_(1))}}` (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".

Part b

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

`t=\frac{5.8-6.9}{\sqrt{(1.2^2)/(35)+(2.9^2)/(35)}}}=-2.074`

P-value

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

`df=n_(1)+n_(2)-2=35+35-2=68`

Since is a one sided test the p value would be:

`p_v =P(t_((68))<-2.074)=0.0209`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we have enough evidence to reject the null hypothesis and then we can conclude that the mean for the group with the vitamin C is lower than the mean for the group without vitamin C