For this assignment, use data from W1 Assignment 3. This week, you will explore the hypothesis that a relationship exists between the misinformation effect (the type of information relayed) and the ac

Week 7 Project

Geanine Patterson

Paired t-test

Descriptive Statistics

Sample

Mean

StDev

SE Mean

Consistent

36

101.611

27.142

4.524

Inconsistent

36

99.194

25.700

4.283

Estimation for Paired Difference

Mean

StDev

SE Mean

95% CI for μd

2.4167

1.9475

0.3246

(1.7577, 3.0756)

μd: mean of (Consistent - Inconsistent)

Test

Null hypothesis

H₀: μd = 0

Alternative hypothesis

H₁: μd ≠ 0

T-Value

P-Value

7.45

<0.0001

Descriptive Statistics

Sample

Mean

StDev

SE Mean

Recall1

72

7.5278

3.2326

0.3810

Recall3

72

5.1250

2.5671

0.3025

Estimation for Paired Difference

Mean

StDev

SE Mean

95% CI for μd

2.4028

1.3179

0.1553

(2.0931, 2.7125)

μd: mean of (Recall1 - Recall3)

Test

Null hypothesis

H₀: μd = 0

Alternative hypothesis

H₁: μd ≠ 0

T-Value

P-Value

15.47

<0.0001

.

In this analysis n1 = 36, sample mean1 = 101.61, sample standard deviation 1 = 27.14,

n2 = 36, sample mean 2 = 99.19, Sample standard deviation 2 = 25.7.

Null Hypothesis (Ho): µ1 = µ2

Alternative Hypothesis (Ha): µ1 ≠ µ2

µ1 = population mean confidence levels for consistent feedback on the car color and µ2 = population mean confidence levels for inconsistent feedback on the car color.

Level of Significance = 0.05

Lower Critical value at 0.05 with n1+n2-2 = 36+36-2 = 72-2 = 70 DF is plus or minus 1.994

Reject Ho if test statistics value is smaller than the smallest critical value or if test statistics value is larger than the upper critical value.

F = (Variance of Sample 1)/ (Variance of Sample 2) = (27.14*27.14)/(25.7*25.7) = 1.11

Upper Critical value at 5% with (36,36) df = 1.75

Since the test statistic values are less than the upper critical value, we can conclude the variance is equal for both the samples. And we will use independent sample t-test for an equal variance to test the original test statistics.

The test statistics value is 0.388470098 which is smaller than 1.994 and more significant than -1.994, so we will not reject the null hypothesis. There is not much difference in the population mean or confidence levels for consistent feedback on car color as well as inconsistent feedback on car color.

Week 8 Project

Geanine Patterson

ANOVA



Method

Null hypothesis

All means are equal

Alternative hypothesis

At least one mean is different

Significance level

α = 0.05

Equal variances were assumed for the analysis.

Factor Information

Factor

Levels

Values

stress level

High, Low, Medium

Analysis of Variance

Source

DF

Adj SS

Adj MS

F-Value

P-Value

StressLevel

261.361111

130.680556

18.76

<0.0001

Error

69

480.583333

6.964976

Total

71

741.944444

Model Summary

R-sq

R-sq(adj)

R-sq(pred)

2.63912407

35.23%

33.35%

29.47%

For this assignment, use data from W1 Assignment 3. This week, you will explore the hypothesis that a relationship exists between the misinformation effect (the type of information relayed) and the ac 1

Null Hypothesis, There is no difference in the population mean recall one regarding three stress-level conditions (low, medium, and high).

Alternative Hypothesis There is a significant difference in the population mean recall one regarding three stress-level conditions (low, medium, and high). At least two groups are significantly different.

µ1 = population mean recall 1 for low stress-level

µ2 = population mean recall 1 for medium stress-level

µ3 = population mean recall 1 for high stress-level

Level of significance = 0.05

Since we have more than two groups, we will use one-way analysis of variance to test the hypothesis about equality of means

The one way ANOVA is used to test the population means recall one concerning three stress-level conditions (low, medium, and high). Population mean for recall 1 differed significantly across the three stress-levels, F (2, 69) = 18.76, p = .0001. Tukey post-hoc comparisons of three groups indicate that the high-level stress group (M = 5.208) gave significantly lesser mean recall at the time.’

We conclude that there is a significant difference in the population mean recall at Time 1 concerning three stress-level conditions (low, medium, and high). Further, from pair-wise p- values table (Multiple Comparisons), we also notice that the p-value is < 0.05 for the two pairs High-medium and high-low.

Week 9 project

Geanine Patterson

Stress and Confidence levels

Correlation

Pearson correlation of StressScore and Confidence = -0.705733

P-Value = <0.0001

The analysis was testing the correlation between the high stress levels and low levels of confidence. In the correlation testing the r – score is -0.707533 and the p-value <0.0001. The results show us a significant relation between stress levels and confidence levels. For an example, if a business man/woman are trying to impress a new client with a product. When high stress levels are present with low confidence levels They would have a less likelihood of the presentation impressing the client.