Assignment 5B
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Assignment 5B
1. Why is the following statement not a proper null hypothesis?
H
o
: M = M
0
a) Correct this null hypothesis.
H
o
: μ = μ
0
b) State the appropriate non
directional
‐
(two tailed)
‐
alternative
hypothesis.
H
1
: μ
μ
0
μ > μ
0 or μ < μ
0
2. When might it be justifiable for a researcher to conclude from a non
‐
significant hypothesis test decision that a) there is no effect, and b) the effect is likely too small to be important?
a. Is justifiable for a researcher to conclude that there is no effect in a non-
significant hypothesis test, where we fail to reject the null hypothesis, it suggests
that the observed data isn't significantly different from what's expected under the
null hypothesis. If the effect size is minimal or close to 0, it may be reasonable for
a researcher to conclude there's no meaningful effect.
b. When there's insufficient evidence to suggest an effect, often indicated by
failing to reject the null hypothesis, the observed effect may be too small to be
practically meaningful.
3. Is a statistical result more likely to lead to rejecting a null hypothesis (i.e., be
statistically significant) at a .01 alpha level or at a .05 alpha level? A more
“conservative” statistical test is one in which it is harder to reject a null
hypothesis. Which alpha level is more conservative, .05 or .01?
An alpha level of .01 is more conservative compared to .05. Making it harder to
reject a null hypothesis at alpha .01 then .05. An alpha of .01 requires more
evidence against the null hypothesis to rejct it than an alpha of .05.
4. Suppose you choose an alpha level of .01 for your null hypothesis significance test (NHST). What is the probability that you will make a Type I error: a) if the null hypothesis is true, and b) if the null hypothesis is false?
a. The probability of making a Type I error is .01. b. If the null hypothesis is false then there is no way to commit Type I error because Type I error can only happen when a null hypothesis is true and rejected. 5. How do the Type I and Type II error rates differ between one- and two
tailed ‐
hypothesis tests?
6. IQ scores among college students are thought to be nearly normally distributed
with mean 116; the standard deviation of the population of college student IQs is not known. UW President Cauce believes that IQ scores of UW students are higher than for the average college student. She has asked you to obtain a random sample of UW students, measure their IQs, and report back to her. Your data from 100 students yield the following:
M = 124.2 s = 12.8
You know that this sample mean, by itself, is not sufficient evidence to support President Cauce’s claim about UW students generally, so you decide to carry out a
null hypothesis significance test. As a researcher, President Cauce commonly used
a .01 alpha level, so you will, too. a) What is your null hypothesis? What is your alternative hypothesis? (Be sure to use useful subscripts!)
H
o
: μ = 116
H
1
: μ
116
μ > 116
or μ < 116
b) Show your computation of the test statistic.
t
obt
=
124.2
−
116
12.8
√
100
=
6.406
c) Determine either a two
tailed
‐
p value
‐
for that test statistic or a critical value.
t
(
df
=
99
,α
=
0.01
,
2
tail
)
=
±
2.626
|
6.406
|
<
2.626
We reject the null hypothesis. d) Can you reject the null hypothesis of no difference? Why or why not?
We can reject the null hypothesis of no difference the computed test statistic 6.406 is far beyond the critical value of 2.626 for two-tailed test.
e) Can you conclude that President Cauce’s belief is supported? Explain.
Yes, President Cauce’s belief is supported since we can reject the null hypothesis UW students IQ scores are equal to average college students. T
obt
of 6.406 is significantly greater than the critical value of 2.626 for a two-tailed test. IQ of UW students (124.2) is higher than the population mean, this supports President Cauce's belief that UW students have higher IQ scores on average compared to the average college student.
7. Return to question 6. If the sample size were actually n = 16 (M = 124.2, s = 12.8), what will be the numerical value of the test statistic? How will your answers
to c), d), and e) change?
t
obt
=
124.2
−
116
12.8
√
16
=
2.563
c .
(
df
=
15
,α
=
0.01
,
2
tail
)
=
±
2.947
|
2.563
|
<
2.947
Fail to reject the null hypothesis. This doesn’t support president Cauce’s belief.. There is not enough evidence to conclude that there is any difference between the IQ scores of UW student to the average college students.
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8. Return again to question 6. If the sample mean was actually found to be M = 118.1 (n = 100, s = 12.8), what will be the numerical value of the test statistic? How will your answers to c), d), and e) change?
t
obt
=
118.1
−
116
12.8
√
100
=
1.641
(
df
=
99
,α
=
0.01
,
2
tail
)
=
±
2.626
|
1.641
|
<
2.626
Fail to reject the null hypothesis. This would not support president Cause’s idea. There is not enough evidence to conclude that UW students IQ scores and the average college students are different. 9.
T
F
A 99% CI for data in question 6 will cover 116.3
T critical = ±
2.626
99%
=
124.2
−
2.626
(
12.8
√
100
)
≤ μ≤
124.2
+
2.626
(
12.8
√
100
)
120.839
≤μ≤
127.561
10.
T
F
A 99% CI for data in question 7 will cover 116.
T critical = ±
2.947
99%
=
124.2
−
2.947
(
12.8
√
16
)
≤μ≤
124.2
+
2.947
(
12.8
√
16
)
114.770
≤ μ≤
133.631
11.
T
F
A 99% CI for data in question 8 will cover 116.
99%
=
118.1
−
2.626
(
12.8
√
100
)
≤ μ≤
118.1
+
2.626
(
12.8
√
100
)
114.739
≤μ≤
121.461
(Be sure you could support your answers to 9, 10, and 11!)
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Note the variable Wplushalf X = W +0.5X
Call:
1m (formula = Y ¯ X + Z + WplushalfX)
Residuals:
Min
10 Median
30
-1.8984 -0.4832 0.2532 0.5756 2.0092
Coefficients:
(Intercept) 4.67189
-0.63364
Z
-2.51074
WplushalfX 11.01191
Max
Estimate Std. Error t value
---
1.66765 2.801
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Pr (>ltl)
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0.1180
<2e-16 ***
<20-16 ***
Signif. codes: 0 *** 0.001 ** 0.01 0.05. 0.1 C
Residual standard error: 1.016 on 24 degrees of freedom
Multiple R-squared: 0.9999, Adjusted R-squared: 0.9999
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e Georgia'.
DEL Element...
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A specific hypothesis where the claim is that the population parameter is equal to 0.
O Any hypothesis that differs from the original claim being made.
O A working hypothesis making a claim about the population parameter in question.
A specific hypothesis where the claim is that the population parameter does not equal 0.
(b) What is an alternate hypothesis H,?
O A working hypothesis making a claim about the population parameter in question.
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O Any hypothesis that differs from the original claim being made.
O A specific hypothesis where the claim is that the population parameter is equal to 0.
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Table 1: Two-way Table of Counts
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Exposure Level
Yes
No
Total
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60
30-89 WLM
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60
90-120 WLM
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A null hypothesis is as follows. Determine the alternative hypothesis.
Question 29
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