# Cretae a Program Code R Coding Task

I have Attached Files: DATATAB_5_2.csv (336 B)This assignment can be submitted on-line here on Blackboard or in class on paper. In any case, R code and R output should be printed out/copied over, and not hand written.1) Page 291, Chapter 6, Excercise #2 (about “Sixty older homes…”).     Note that the F table starts on page 689 or you can use an F calculator such as the one at https://homepage.divms.uiowa.edu/~mbognar/applets/f.html2) The data for Table 5.2 in Section 5.2.6 on page 209 is attached as a .csv file, or can be found at https://people.stat.sc.edu/habing/courses/DATATAB_5_2.csv .   If you are using the R console it should be read in using the read.csv function instead of read.table (the function works the same as read.table as far as how the file name goes in and using header=T).   If you read the file in as tab52, for example, then attach(tab52) would create variables called location and heights. Use R to find the top two rows of the ANOVA table for this problem.  Verify that the MSE (called Mean Sq Residuals in R) is the same as the sp2 value on page 210, and that the F value you found for the ANOVA is very close to the value of the t-squared  for the t found on page 210. we are using the 4th addition from statical method by william j , find the attachment for page 210

PDF
*Statistical Methods, 4th Edition
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The computed t statistic is
1.6860 – 1.3215
t
5.0495 + 3.5175
1 1
+
20 26
44
0.3645
70.1947)(0.08846)
0.3654
0.1312
= 2.778.
We have decided that a significance level of 0.01 would be appropriate. For this test we need the t
distribution for 20 + 26–2 = 44 degrees of freedom. Because Appendix Table A.2 does not have entries
for 44 degrees of freedom, we use the next smallest degrees of freedom, which is 40. This provides for a
more conservative test; that is, the true value of a will be somewhat less than the specified 0.01.
(Statistical software, Excel, and most scientific calculators will give you the exact critical value.) Using this
approximation, we see that the rejection region consists of absolute values exceeding 2.7045.
The value of the test statistic exceeds 2.7045 so the null hypothesis is rejected, and we determine
that the average heights of plants differ between the two locations. Using readily available Microsoft
Excel, or TI-84/89 calculators, the exact p value for the test statistic is 0.008.
The 0.99 confidence interval on the difference in population means, (uly – M2), is
51-52 Eta/2V/(1/ni + 1/12),
which produces the values
0.3645 + 2.7045(0.1312) or 0.3645 +0.3548,
which defines the interval from 0.0097 to 0.7193. The interval does not contain zero, which agrees with
the results of the hypothesis test.
5.2.7 Variances Unknown but Not Equal
The adaptation for the case where the variances differ is well known and presented
below. However, we stress that whenever possible, we first should search for a trans-
formation of the data that will make the sample variances more nearly equal. This is
because it makes more sense to compare the locations of the two distributions if they
are on similar scales. Consider the distributions in Figure 5.2, where the apparent
dif-
ference in the means is about 3.0. But is 3.0 a big difference or a small one? In terms
of the “narrow” distribution, 3.0 is quite a big difference, but in terms of the wider
distribution, it is only a moderate difference.
Transformations are particularly helpful when the magnitude of the variance is
systematically related to the size of the mean. For example, for many biological organ-
isms, populations with larger means also have larger variances. Analyzing transformed
data, such as log y rather than y, can equalize the variance and reduce skewness. If so,
N
W
ENG
5:14 PM
2/23/2022
PDF
*Statistical Methods, 4th Edition
Х
+

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0 File | C:/Users/alyah/Desktop/Stat2/Statistical%20Methods,%204th%20Edition%20by%20Donna%20L.%20Mohr%2…
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Question 1 of the E…
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11111
Total
2. Sixty older homes are randomly divided into four groups. Homes in each
group are retrofitted with a different type of insulation, and then the change in
energy consumption is recorded. A portion of the ANOVA table is given below.
Complete the table, and draw the appropriate conclusion assuming a = 5%.
=
Source
df
SS
MS
F
Between
Within
Total
155
896
3. Thirty participants are randomly divided into three groups. Each participant is
given a list of words to memorize. The groups differ in the kinds of words given
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w
ENG
5:25 PM
2/23/2022
1
PDF
Statistical Methods, 4th Edition b X
+

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(+1
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Question 1 of the E…
>
ili
229
of 768
Q
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CD
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T

o
3g
210
Statistical Methods
The computed t statistic is
1.6860 – 1.3215
1
5.0495 +3.5175
44
+
1
26
20
0.3645
(0.1947)(0.08846)
0.3654
0.1312
= 2.778.
We have decided that a significance level of 0.01 would be appropriate. For this test we need the t
distribution for 20 + 26-2 = 44 degrees of freedom. Because Appendix Table A.2 does not have entries
for 44 degrees of freedom, we use the next smallest degrees of freedom, which is 40. This provides for a
more conservative test; that is, the true value of a will be somewhat less than the specified 0.01.
(Statistical software, Excel, and most scientific calculators will give you the exact critical value.) Using this
approximation, we see that the rejection region consists of absolute values exceeding 2.7045.
The value of the test statistic exceeds 2.7045 so the null hypothesis is rejected, and we determine
that the average heights of plants differ between the two locations. Using readily available Microsoft
Excel, or TI-84/89 calculators, the exact p value for the test statistic is 0.008.
The 0.99 confidence interval on the difference in population means, (, – M2), is
51-72 Eta/2V/(1/ni + 1/n2),
which produces the values
0.3645 2.7045(0.1312) or 0.3645 + 0.3548,
which defines the interval from 0.0097 to 0.7193. The interval does not contain zero, which agrees with
the results of the hypothesis test.
5.2.7 Variances Unknown but Not Equal
The adaptation for the case where the variances differ is well known and presented
below. However, we stress that whenever possible, we first should search for a trans-
formation of the data that will make the sample variances more nearly equal. This is
because it makes more sense to compare the locations of the two distributions if they
are on similar scales. Consider the distributions in Figure 5.2, where the apparent dif-
ference in the means is about 3.0. But is 3.0 a big difference or a small one? In terms
of the “narrow” distribution, 3.0 is quite a big difference, but in terms of the wider
distribution, it is only a moderate difference.
Transformations are particularly helpful when the magnitude of the variance is
systematically related to the size of the mean. For example, for many biological organ-
isms, populations with larger means also have larger variances. Analyzing transformed
data, such as log y rather than y, can equalize the variance and reduce skewness. If so,
N
W
ENG
6:26 PM
2/23/2022

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