# Answers to Practice Final Part 1 ```I’m sorry about all tests being fail to reject.
I did double check these, so I don’t think I made a mistake, but it’s always possible. Let
me know. And the drawings would have taken way too much time and not been helpful,
so I didn’t do those.
1)
H0: &micro; = 60
H1: &micro; ≠ 60
Assumptions: sample must be large enough (&gt;30) to use CLT or the data is distributed
normally.
Distribution: Normal (z test) distribution. Two-tail. Degrees of freedom: NA
Critical values: &plusmn;1.96
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though
it is true.
test statistic = -.981
Since the test statistic (-.981) is not more extreme than the critical value (-1.96), we fail to
reject the null hypothesis
Since the p-value (.326) is greater than the level of significance (α=.05), we fail to reject
the null hypothesis
In English: there is insufficient evidence to conclude that the mean reading score for this
fifth grade class is different than the national average.
2)
H0: σ = 15
H1: σ ≠ 15
Assumptions: data is distributed normally.
Distribution: χ2. Two-tail. Degrees of freedom: n-1 = 11
Critical values: 3.82, 21.92
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though
it is true.
test statistic = (11 * 16.8372) / 152 = 13.86
Since the test statistic (13.86) is not more extreme than the critical values (3.82 or 21.92),
we fail to reject the null hypothesis
No p-value
In English: there is insufficient evidence to conclude that the standard deviation of reading
scores for this fifth grade class is different than the national average.
3)
H0: &micro;O = &micro;N
H1: &micro;O &lt; &micro;N
Assumptions: samples (both) must be large enough (&gt;30) to use CLT or the data is
distributed normally.
Distribution: Note: we might do this as a Z-test by using the national standard deviation.
But I’m going to assume that the standard deviations of the new program may not be
known, so it’s going to be a t-test.
t-distribution. One-tail. Since I’ve got the old first (on the left above and will be list 1),
this is a left tail test. Degrees of freedom: (from calculator) 21
Critical values: -1.72
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though
it is true.
test statistic = -1.11
Since the test statistic (-1.11) is not more extreme than the critical value (-1.72), we fail to
reject the null hypothesis
Since the p-value (.139) is greater than the level of significance (α=.05), we fail to reject
the null hypothesis
In English: there is insufficient evidence to conclude that the mean reading score for the
new teaching program are lower than the older program.
4)
H0: σO = σN
H1: σO &lt; σN
Assumptions: the data is distributed normally.
Distribution: F-test. One-tail. Since I’ve got the old variant first (on the left, list #1 in the
calculator), this will be the numerator we’re testing if it is bigger than the New which will
be the denominator. So the F-stat should be &gt;1 which makes this a right tail test. Degrees
of freedom: 11 &amp; 11.
Critical values: 2.82
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though
it is true.
test statistic = 2.61
Since the test statistic (2.61) is not more extreme than the critical value (2.82), we fail to
reject the null hypothesis
Since the p-value (.064) is greater than the level of significance (α=.05), we fail to reject
the null hypothesis
In English: there is insufficient evidence to conclude that the mean reading score for the
new variant of the reading program has a lower standard deviation than the old variant of
the new program.
5)
H0: &micro;1 = &micro;2 = &micro;3
H1: at least one mean (&micro;) is different than the others.
Assumptions: data are distributed normally, independent, with the same standard deviation.
Distribution: F-test. One-tail. Right tail. Degrees of freedom: 2 &amp; 33
Critical values: 3.28
α = .05 which means there’s a 5% probability of rejecting the null hypothesis even though
it is true.
test statistic = 1.01
Since the test statistic (1.01) is not more extreme than the critical value (3.28), we fail to
reject the null hypothesis
Since the p-value (.373) is greater than the level of significance (α=.05), we fail to reject
the null hypothesis
In English: there is insufficient evidence to conclude that at least one mean reading score is
different than the others.
```