Week 1 Project

Running head: WEEK 1 PROJECT 1
Week 1 Project
Name
Institution

WEEK 1 PROJECT 2
Week 1 Project
Part 1
The first step involves checking the data for normality in order to decide whether to use
parametric or non-parametric test to compare the mean confidence levels for consistent and
inconsistent feedback on car color. The figure 1 shows the normality plot for consistent response.
The figure 1 also includes Ryan-Joiner (Similar to Shapiro) test results. The hypothesis for the
null hypothesis for RJ is the data is from a normally distributed population against non-normal
data (Gio & Caraka, 2019).
1751501251007550
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean 101.6
StDev 27.14
N 36
RJ 0.989
P-Value >0.100
Consistent
Percent
Figure 1: Probability Plot of Consistent and Test
Normal
Majority of the plots are on or close to the diagonal line. Further the RJ = 0.989 with a
corresponding p-value = 0100 greater than alpha = 0.05, therefore at 95% significance level fail
to reject the null hypothesis and conclude that the data is from a normally distributed population.
Similarly figure 2 shows the normality plot and RJ test results for inconsistent variable.

WEEK 1 PROJECT 3
1751501251007550
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean 99.19
StDev 25.70
N 36
RJ 0.989
P-Value >0.100
Inconsistent
Percent
Figure 2: Probability Plot of Inconsistent and RJ test
From figure 2 majority of the plots are on or close to the diagonal line. Further the RJ =
0.989 with a corresponding p-value = 0.100 greater than alpha = 0.05, therefore at 95%
significance level fail to reject the null hypothesis and conclude that the data is from a normally
distributed population. Now that the normality assumption is confirmed the test for comparing
the confidence levels is two sample student t-test with the hypothesis (Lesik, 2018):
H0: The μ (Consistent) - μ (Inconsistent) = 0
Ha: The μ (Consistent) - μ (Inconsistent) ≠ 0
Test statics p-value for two sample t-test. Decision, if the p-value is less than α = 0.05
reject null hypothesis of no difference. The calculated t-value = 0.39 with corresponding p-value
= 0.699 degrees of freedom = 69. The p-value = 0.699 > α = 0.05, fail to reject null hypothesis
and conclude that at 95% significance level there is no difference on the confidence levels of
participants given consistent feedback on car color and those given inconsistent feedback on car
color. The average confidence level for percentages of those who were consistent in response to