The average house has 14 paintings on its walls. Is the mean different for houses owned by teachers? The data show the results of a survey of 13 teachers who were asked how many paintings they have in their houses. Assume that the distribution of the population is normal.
15, 13, 14, 15, 13, 11, 13, 12, 15, 14, 11, 13, 12
What can be concluded at the αα = 0.10 level of significance?
Solution(a)
In this question we will use t test
Solution(b)
Given in the question
Null hypothesis H0:
= 14
Alternative hypothesis Ha:
14
No. of sample = 13
Sample mean = (15+13+14+15+13+11+13+12+15+14+11+13+12)/13 = 171/13
= 13.15
Sample standard deviation can be calculated as
Sample standard deviation =
sqrt((Summation(Xi-mean)^2)/(N-1))
X |
Xi-mean |
(Xi-mean)^2 |
15 |
1.85 |
3.4225 |
13 |
-0.15 |
0.0225 |
14 |
0.85 |
0.7225 |
15 |
1.85 |
3.4225 |
13 |
-0.15 |
0.0225 |
11 |
-2.15 |
4.6225 |
13 |
-0.15 |
0.0225 |
12 |
-1.15 |
1.3225 |
15 |
1.85 |
3.4225 |
14 |
0.85 |
0.7225 |
11 |
-2.15 |
4.6225 |
13 |
-0.15 |
0.0225 |
12 |
-1.15 |
1.3225 |
Summation(Xi-mean)^2 |
23.6925 |
Standard deviation = sqrt(23.6925/12) = 1.41
Solution(c)
test statistic can be calculated as
Test stat = (Xbar -
)/S/sqrt(n) = (13.15-14)/1.41/sqrt(13) = -2.17
Solution(d)
From t table we found p-value at df = 12 and this is two tailed
test 0.0507
P-value = 0.0507
Solution(e)
Here alpha =0.10
So P-value is less than alpha vlaue i.e. (0.0507<0.10)
Solution(f)
Based on this we will reject Null hypothesis.
Solution(g)
Its correct answer is C. i.e. the data suggest the population mean
is significantly different from 14 at alpha = 0.10, so there is
sufficient evidence to conclude that the population mean number of
painting that are in teachers houses is different from 14.
Solution(h)
Its answer is D i.e. there is a 5.07% chance that the population
mean number of paintings that are in teacher's houses is not equal
to 14.
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