Question

The following data represent the results from an independent-measures study comparing two treatment conditions.

Treatment
One Treatment
Two
4.4 2.7
4.3 2.8
4.3 1.5
5.1 5
3.8 5
5.3 4.1
5.7 3.5
4.5 3


Conduct ANOVA with
for this data and calculate the
-ratio and
-value. Round your answer to three decimal places. Assume all population and ANOVA requirements are met.

F-ratio:

p-value:


Now, conduct a t-test with
on the same data and calculate the
-statisitc and
-value. Round your answer to three decimal places.

t-statistic:

p-value:


Observing the p-values, was there a difference in the result of the hypothesis test?
Yes, the p-values are different enough to change the outcome of the test.
No, the p-values aren't that far off where it would change the outcome of the test.

Answer 1

Explanation:

To conduct ANOVA, we first need to calculate the sum of squares (SS) for each factor and for error.

SS(total) = SS(Treatment) + SS(Error)

We can calculate the means and sum of squares for each treatment group:

Treatment One: Mean = 4.5, SS = 3.15

Treatment Two: Mean = 3.43, SS = 5.727

Next, we can calculate the grand mean:

Grand Mean = (4.5 + 3.43) / 2 = 3.965

Using the grand mean, we can calculate the total sum of squares:

SS(total) = ∑(X-Grand Mean)^2 = 17.7175

Using the sum of squares for each treatment group and the total sum of squares, we can calculate the sum of squares for error:

SS(Error) = SS(total) - SS(Treatment) = 9.335

Next, we can calculate the degrees of freedom:

df(Treatment) = k - 1 = 2 - 1 = 1

df(Error) = N - k = 8 - 2 = 6

df(total) = N - 1 = 8 - 1 = 7

Now we can calculate the mean squares:

MS(Treatment) = SS(Treatment) / df(Treatment) = 3.15 / 1 = 3.15

MS(Error) = SS(Error) / df(Error) = 9.335 / 6 = 1.556

Using the mean squares, we can calculate the F-ratio:

F-ratio = MS(Treatment) / MS(Error) = 3.15 / 1.556 = 2.022

To find the p-value, we need to compare the F-ratio to the F-distribution with (1, 6) degrees of freedom. Using a significance level of 0.05, we find that the critical F-value is 5.143. Since our F-ratio is less than the critical F-value, we fail to reject the null hypothesis.

Therefore, the results of the ANOVA suggest that there is not enough evidence to conclude that the means of the two treatment groups are significantly different.

Now we can conduct a t-test to compare the means of the two groups:

t = (mean1 - mean2) / (s / sqrt(n))

where mean1 and mean2 are the sample means, s is the pooled standard deviation, and n is the sample size.

Using the given data, we can calculate the sample means and standard deviations for each group:

Treatment One: Mean = 4.5, s = 0.72

Treatment Two: Mean = 3.43, s = 1.45

Using the pooled standard deviation formula, we can calculate the pooled standard deviation:

s = sqrt(((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2)) = sqrt(((7 * 0.72^2) + (7 * 1.45^2)) / (14 - 2)) = 1.074

Using the means, pooled standard deviation, and sample sizes, we can calculate the t-statistic:

t = (4.5 - 3.43) / (1.074 / sqrt(7)) = 2.046

To find the p-value, we need to compare the t-statistic to the t-distribution with (6) degrees of freedom. Using a significance level of 0.05 and a two-tailed test, we find that the critical t-value is approximately