Description
- Select at least three variables that you believe have a linear relationship.
- Specify how you will measure each of these variables (i.e., what instrument will you use and provide an APA reference for the instrument)
- Collect the data for these variables and describe the data collection technique and why it was appropriate as well as why the sample size was best.
- Submit the data collected by submitting the SPSS data file with your submission.
- Find the Correlation coefficient for each of the possible pairings of variables and describe the relationship in terms of strength and direction.
- Find a linear model of the relationship between the three (or more) variables of interest. Identify the predictor variables and the criterion variable.
- Provide an output of the SPSS results and interpret the results using correct APA style.
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Explanation & Answer
Attached.
Chapter 13
13.6
(a)
Hypotheses are;
Null hypothesis, 𝐻0 : 𝜇 = 32
Alternative hypothesis, 𝐻0 : 𝜇 ≠ 32
Test statistics, t =
𝑥̅ −µ
𝑠
√𝑛
Where;
𝑥̅ , sample mean = 34.89
µ, population mean = 32
s, standard deviation = 3.02
n, number of rats in the sample = 7
=
34.89−32
3.02
√7
= 2.532
Degrees of freedom, 𝑑𝑓 = 𝑛 − 1 = 7 − 1 = 6
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 𝑃(𝑡6 < 2.532) = 0.0223
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = (2 × 0.0223) = 0.0446
Since the p-value < 0.05, reject the null hypothesis. This simply implies that the shock
treatment has caused a change in the results of the water maze study.
(b)
95% confidence interval is;
𝑥̅ = 34.89
Margin of error = 1.96
Standard deviation = 3.02
Confidence Interval = 34.89 ± (1.96 × 3.02)
=34.89 ± 5.9192
Confidence Interval = (34.89 - 5.9192, 34.89 + 5.9192)
Confidence Interval = (28.97, 40.81)
(c)
We have 95% confidence that the number of trials required is between 28.97 and 40.81
13.8
(a)
Hypotheses are:
Null hypothesis, 𝐻0 : 𝜇 = 90
Alternative hypothesis, 𝐻1 : 𝜇 ≠ 90
Test statistic; t
=
𝑥̅ −µ
𝑠
√𝑛
Where;
𝑥̅ , sample mean = 88
µ, population mean = 90
s, standard deviation = 9
n, number of rats in the sample = 28
t=
88−90
9
√28
= -1.18
Degrees of freedom = 𝑑𝑓 = 𝑛 − 1 = 28 − 1 = 27
For 27 degrees of freedom;
p - value = 2 × P(t < -1.18) = 0.2499
Since the p - value > 0.05, we do not reject the null hypothesis since there we do not have
sufficient evidence to conclude that drinking coffee just before going to sleep affects the
amount of dreamtime.
(b)
Since we are not rejecting the null hypothesis, as the question suggests, we do not need to
construct the confidence interval.
13.9
(a)
In the first scenario where I am a car manufacturer, I would prefer a smaller sample size as the
conclusions arrived at here would relate to a smaller population and therefore, less underlying
issues.
(b)
In this case, I would prefer a much larger sample size as this way, there will be more variability
of flaws and other mandated issues that would need to be addressed.
13.10
a)
Here level of significance is large because here, z is not appropriate.
b)
The true critical value is larger than z because the level of significance is large.
Chapter 14
14.11
Mean
Standard deviation
Committee
2
5
20
15
4
10
9.3333333
7.03325434
Solitary
3
8
7
10
14
0
7.0
4.97996
Now we will use the above-given data for Hypothesis testing.
Null hypothesis,𝐻0 : 𝜇1 − 𝜇2 = 0
Alternative hypothesis, 𝐻1 : 𝜇1 − 𝜇2 ≠ 0
𝑠2
Standard Error, 𝑆𝐸 = √(𝑛1
1
2
2
7.033
4.980
𝑆𝐸 = √ 6 + 6
Test statistic, t =
𝑡=
𝑠2
+ 𝑛2 )
2
= 3.518
𝑥̅ 1 −𝑥̅ 2
𝑆𝐸
9.333−7.000
= 0.663
3.518
The corresponding rejection region will be 𝑡 < �...