Question

Low	Medium	High
14	18	16
6	17	20
11	23	18
12	17	16
11	10	15
9	20	16
6	15	16
12	17	17
14	25	11
11	11	13
4	18	22
9	18	16
7	20	10
12	21	18
12	27	9
17	13	19
11	22	16
8	17	12

Our worksheets lists the number of 3+ syllable words in each magazine grouped by education level. Use your spreadsheet program to conduct a one-way ANOVA on the data. You can assume the data meets all the assumptions required for a one-way ANOVA (normally distributed, no outliers, etc) What is your critical cutoff value for F?

Which of the following would be an appropriate p value to report for this test?

Should we:

		fail to reject the null hypothesis - magazine type had no effect on the number of 3+ syllable words in the ads.
		reject the null hypothesis - the number of 3+ syllable words was significantly different in at least one of the magazine types.

Let's do post-hoc comparisions between all the groups. Use t-tests to contrast the low-medium, medium-high, and low-high groups.

Before you conduct your t-tests you will need to use the "F-Test Two-Sample for Variances" tool. According to the results from this tool, how many of your contrasts will use the t-test that assumes equal variances?

What was the t Stat for your Low-Medium contrast?

What was the t Stat for your Medium-High contrast?

What is the t Stat for your Low-High contrast?

To lower our chance of making a type I error we should use Bonferroni's correction to adjust our alpha level. Instead of rejecting the null hypothesis if p < .05, what will be our new requirement?

		p < .03
		p < .025
		p < .01
		p < .017

Answer 1

Here

F critical value = 3.1788

F > F critical

Reject the null hypothesis - the number of 3+ syllable words was significantly different in at least one of the magazine types.

F-Test

H₀: σ²₁ = σ²₂   (there is no difference between variances)

H₁: σ²₁ ≠ σ²₂   (there is a difference between variances)

T-Test

For all Low-Medium, Medium-High, High-Low the null and alternate hypothesis will be

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)

H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

To lower our chance of making a type I error we should use Bonferroni's correction to adjust our alpha level. Instead of rejecting the null hypothesis if p < .05

Here,there are three comparisons

using Bonferroni's correction, we should get 0.05/3=0.0167 and hence all that should sum up to 0.05.

P<0.017