Question 1087882: A clinical psychologist has noted that autistic children seem to respond to treatment better if they are in a familiar environment. To evaluate the influence of environment, the psychologist selects a group of 15 autistic children who are currently in treatment and randomly divides them into three groups. One group continues to receive treatment in the clinic as usual. For the second group, treatment sessions are conducted entirely in the child's home. The third group gets half of the treatment in the clinic and half at home. After 6 weeks, the psychologist evaluates the progress for each child. The data are as follows:
Clinic Home Both
1 2 4
1 7 1
5 2 2
2 4 2
1 5 6 N = 15
M = 2 M = 4 M = 3 G = 45
SS = 12 SS = 18 SS = 16 Ex2 = 191
Do the data indicate any significant differences between the three settings? Test at the .05 level of significance. Be sure to show all formulas with symbols (and plug in numbers), steps, processes and calculations for all parts of the answers.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let,
 = population mean score of the "clinic" group
 = population mean score of the "home" group
 = population mean score of the "both" group
 is the greek letter mu
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Null Hypothesis
H0:  (all the population means are the same)
Alternative Hypothesis:
H1:  or  or  (at least one of the population means is different)
We're going to perform a one-way ANOVA test to see if we reject the null hypothesis or not.
We will reject the null hypothesis if the p value is less than the significance level alpha = 0.05
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Given Data Table:
From the table we're given
xbar1 = 2
xbar2 = 4
xbar3 = 3
as shown in cells A7, B7, & C7. I'm going to use xbar instead of M. Note the numbers after each "xbar" to indicate the proper label (eg: xbar1 is the mean for group1)
n1 = number of items in group1 = number of scores in the "clinic" group = 5
n2 = number of items in group2 = number of scores in the "home" group = 5
n3 = number of items in group2 = number of scores in the "both" group = 5
N = number of items total
N = n1+n2+n3
N = 5+5+5
N = 15
k = number of groups (or columns)
k = 3
GM = Grand Mean
GM = (Sum of the xbar values)/(number of groups)
GM = (xbar1+xbar2+xbar3)/k
GM = (2+4+3)/3
GM = 9/3
GM = 3
Use the grand mean (GM), the sample sizes(n1 through n3), and the sample means (xbar1 through xbar3) to compute the value of SS_B
SS_B = Sum of Squares Between groups
SS_B = n1*(xbar1-GM)^2 + n2*(xbar2-GM)^2 + n3*(xbar3-GM)^2
SS_B = 5*(2-3)^2 + 5*(4-3)^2 + 5*(3-3)^2
SS_B = 10 <<<--- This value is shown in the ANOVA table of the "Excel" section (shown below)
From the table (cells A8,B8,C8) we see that
SS1 = 12
SS2 = 18
SS3 = 16
where "SS" means "sum of squares". Specifically it's the sum of the squared deviations from the mean. Like with the xbar values, I'm attaching numbers to the end of SS so I can keep track of the values.
Using the three SS values, we can say
SS_W = Sum of Squares Within groups
SS_W = SS1+SS2+SS3
SS_W = 12+18+16
SS_W = 46 <<<--- This value is shown in the ANOVA table of the "Excel" section (shown below)
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Let's use the SS_B and SS_W values to compute MS_B and MS_W
MS_B = Mean Square Between groups
MS_W = Mean Square Within groups
Recall earlier that
k = 3
N = 15
So,
Degrees of freedom in the numerator = df_N = k-1 = 3-1 = 2
Degrees of freedom in the denominator = df_D = N-k = 15-3 = 12
Allowing us to compute the mean square values
MS_B = (SS_B)/(df_N)
MS_B = (10)/(2)
MS_B = 5 <<<--- This value is shown in the ANOVA table of the "Excel" section (shown below)
MS_W = (SS_W)/(df_D)
MS_W = (46)/(12)
MS_W = 3.8333 <<<--- This value is shown in the ANOVA table of the "Excel" section (shown below)
F = F Test Statistic
F = (MS_B)/(MS_W)
F = (5)/(3.8333)
F = 1.3043591683406
F = 1.3043 <<<--- This value is shown in the ANOVA table of the "Excel" section (shown below)
Now use a calculator to find the p-value.
If you want to use a TI84 calculator (or similar), then follow these steps
Step 1) Clear the home screen
Step 2) Hit the "2ND" key up at the top left (usually the key is blue)
Step 3) Press the "VARS" button which is just below the scroll directional pad
Step 4) Scroll down til you reach "Fcdf(", which is just after item 9. Hit enter once you get to "Fcdf("
Step 5) After "Fcdf(" shows up on the home screen, type in the following: 1.3043,99,2,12). The value 1.3043 is the F test stastistic found earlier above. The value 99 is some large number to help set up the right side of the boundary. The value 2 refers to the df_N value (see above) and the 12 is the df_D value (see above). Don't forget to close off with a parenthesis. Then hit enter. Once you do, you should get something like this
The value that pops up, which is roughly 0.3072 when rounded to 4 decimal places, is the P-value. This value is shown in the ANOVA table of the "Excel" section (shown below)
Because the P-value is larger than alpha = 0.05, this means we fail to reject the null hypothesis (H0).
In other words, we have no choice but to accept the null because we don't have enough statistically significant evidence to overturn the null.
In plain English, we conclude that the three groups have the same average scores (because we concluded that was true)
The question asks "Do the data indicate any significant differences between the three settings?"
The answer here is "No, there isn't enough evidence to indicate there are any differences between the three settings".
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Excel Section:
This section is optional. If you have Microsoft Excel and wish to quickly compute the ANOVA table (to quickly do the hypothesis test and to check your work), then read on. If you don't have Excel, you can use free programs like OpenOffice or LibreOffice to get the job done (though the steps may vary).
The first step is to get all the data into Excel. Make sure that the columns are properly labeled so you can keep track of the data.
After the data is in the spreadsheet, you'll do a data analysis. Go to the "data" tab and click "data analysis" on the far right side. You may need to install the analysis toolpack first. Though I think some versions of Excel have it already built in.
When you click the "data analysis" button, and after you select "ANOVA: Single Factor", it will ask you to input the data. You'll simply select the cells you typed in earlier. Make sure to include the headers and check off "labels in first row" so Excel knows that first row is not data.
If you have your data groups in columns, then check the "columns" box. Otherwise check "rows". I'd stick to columns because that's how your table is initially set up.
Leave alpha as 0.05
You have the option to output the ANOVA table in the same worksheet or on a different worksheet. Let's put it on the next worksheet.
After everything is set up and executed you should get something like this
Note: I rounded everything to 4 decimal places so things don't get too cluttered
Every value in that table was computed earlier. The table gives a neat compact way to represent all of the pertinent values. Ultimately the only value we care about is the P-value. This is shown in cell F12 and it matches up with what the TI calculator got earlier.
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