{"id":148,"date":"2018-03-08T04:05:00","date_gmt":"2018-03-08T04:05:00","guid":{"rendered":"https:\/\/kindsonthegenius.com\/blog\/2018\/03\/08\/one-way-anovaanalysis-of-variance-problem-question-11\/"},"modified":"2020-11-05T14:22:57","modified_gmt":"2020-11-05T13:22:57","slug":"one-way-anovaanalysis-of-variance-problem-question-11","status":"publish","type":"post","link":"https:\/\/kindsonthegenius.com\/blog\/one-way-anovaanalysis-of-variance-problem-question-11\/","title":{"rendered":"One-Way ANOVA(Analysis of Variance) Problem – Question 11"},"content":{"rendered":"
\n
Question 11 <\/b><\/span><\/span>
Using the following three groups of data, perform a one-way analysis of variance using \u03b1  = 0.05.<\/p>\n\n\n\n\n\n\n\n\n
Group 1<\/b><\/td>\nGroup 2<\/b><\/td>\nGroup 3<\/b><\/td>\n<\/tr>\n
51<\/td>\n23<\/td>\n56<\/td>\n<\/tr>\n
45<\/td>\n43<\/td>\n76<\/td>\n<\/tr>\n
33<\/td>\n23<\/td>\n74<\/td>\n<\/tr>\n
45<\/td>\n43<\/td>\n87<\/td>\n<\/tr>\n
67<\/td>\n45<\/td>\n56<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Read about one-way ANOVA and the steps here<\/a>
Watch the video here for step by step explanation<\/a>
Solutions Steps<\/b><\/span><\/span>
As usual, we would transfer the table of data into MS Excel so it becomes easier to work with. The excel table is given below.<\/p>\n

<\/a><\/div>\n

Figure 1.<\/p>\n

Step 1: Calculate all the means<\/b>
This have been done in the excel sheet as you can see in Figure 1. <\/p>\n

Step 2: Set up the Null and Alternate Hypothesis <\/b>
You should already know how to do this by now
H0<\/sub>:<\/b> \u03bc1<\/sub> = \u03bc2<\/sub> = \u03bc3<\/sub>
Ha<\/sub><\/b>: \u03bc1<\/sub> \u2260 \u03bc2<\/sub> \u2260 \u03bc3  <\/sub><\/p>\n

We also specify the  as well as the rejection criteria.
\u03b1 = 0.05
Rejection criteria: K0.05<\/sub> < F
This means that if the critical value of F from tables is less than the calculated value of F, we reject the null hypothesis<\/p>\n

Step 3: Calculate the Sum of Squares<\/b>
The formular for sum of squares are given below<\/p>\n

<\/a><\/div>\n

I have expanded the excel sheet to calculate the values of this terms in the equation
Take some time to study it. The formulas are quite simple and clear. If you have issue though let me know in the comment box to the left of this page<\/p>\n

<\/a><\/div>\n

In the excel sheet above, you can verify that the total sum of squares  equals to the sum of squares between and sum of squares between.<\/p>\n

Step 4: Calculate the Degrees of Freedom<\/b>
The degrees of freedom are calculated using the formula below.
dftotal =        n – 1 = 15 – 1 = 14
dfwithin =     n – k = 15 – 3 = 12
dfbetween = k – 1 = 3 – 1 = 2<\/p>\n

Verify that dftotal = dfbetween + dfwithin<\/p>\n

Step 5: Calculate the Mean Squares<\/b>
The formula is given below for the mean squares between<\/p>\n

<\/a><\/div>\n

The formula for the mean squares within<\/p>\n

<\/a><\/div>\n

<\/span>Step 6: Calculate the F Statistic<\/span><\/b>
Then we go ahead to calculate the F statistic using MSB and MSW<\/p>\n

<\/a><\/div>\n

At this point, our excel sheet have been extended to calculate all the values. The formula used is written beside some of the values<\/p>\n

Our final excel sheet would look like the one below:<\/p>\n

<\/a><\/div>\n

 Get this excel sheet from here<\/a><\/p>\n

Step 7: Look up F from table and state your Conclusion<\/b>
Access table from here <\/a>
From the table of F distribution, the critical value of F for 0.05 significance and degrees of freedom of(df1 = 12 and df2 = 2) we have:<\/p>\n

F = 3.89<\/i><\/span><\/b><\/p>\n

Since the calculated(absolute value) of F is greater than the tabulated value, we reject the null hypothesis and conclude that at least two of the means are significantly different from each other.<\/p>\n

Hurray!!! We are done finally! So go take a bottle of beer! Then take the quiz below<\/p>\n

Watch the video here for step by step explanation<\/a>
Try another ANOVA problem(Question 10)<\/a><\/p>\n

Take a Simple Quiz<\/b>
Complete the Anova table below with actual values from this example
If you do it, let me know in the comment box by the left of this page.<\/p>\n

<\/a><\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 11 Using the following three groups of data, perform a one-way analysis of variance using \u03b1  = 0.05. Group 1 Group 2 Group 3 … <\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0},"categories":[223],"tags":[],"_links":{"self":[{"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/posts\/148"}],"collection":[{"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/comments?post=148"}],"version-history":[{"count":5,"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/posts\/148\/revisions"}],"predecessor-version":[{"id":1772,"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/posts\/148\/revisions\/1772"}],"wp:attachment":[{"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/media?parent=148"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/categories?post=148"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kindsonthegenius.com\/blog\/wp-json\/wp\/v2\/tags?post=148"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}