Issuu on Google+

Assignment #5 – Jamie Hoag For this assignment, you are required to analyze two datasets: 1. You are required to test the hypothesis that individuals receiving the standard type of advertisement, on average, spend significantly less money than individuals receiving the other type of advertisement. Which test will you use to test this hypothesis? Why is this test relevant? Test and report whether the assumptions that underlie this test are satisfied. Report the results using statistical terminology and interpret them. I will be doing an Independent Means T-Test for this scenario. “This test is used when there are two experimental conditions and different participants were assigned to each condition� (Field, 2009, p. 325). In this scenario I must identify that the group receiving standard advertisement will spend significantly less money than the group receiving promotional advertisements, which are two groups of different participants assigned with a different experimental difference. Below are the results I found from testing the Independent Means T-Test. Group Statistics Type of advertisement

N

Mean

Std. Deviation

Std. Error Mean

$ spent during promotional

Standard

250

1566.3890

346.67305

21.92553

period

New Promotion

250

1637.5000

356.70317

22.55989

*It is important to note the Mean and Std. Error Mean column in this table. Independent Samples Test Levene's Test for Equality of

t-test for Equality of Means

Variances F

Sig.

t

df

Sig. (2-

Mean

Std. Error

95% Confidence Interval of

tailed)

Difference

Difference

the Difference Lower

Equal variances $ spent during

assumed

promotional period

Equal variances not

1.190

.276

assumed

*It is important to note the df and t column in this table.

Upper

-2.260

498

.024

-71.11095

31.45914

-132.91995

-9.30196

-2.260

497.595

.024

-71.11095

31.45914

-132.92007

-9.30183


On average, individuals spend more money when there is a new promotion (M = $1637.50, SE = $22.55) compared to a standard advertisement (M = $1566.39, SE = $21.93). This difference was significant t(498) = -2.26, p > 0.05; however, it did represent a small-sized effect r = 0.10 (p. 341). The hypothesis was supported by the data, which shows that customers who see a new promotion will spend more money, than if they receive a standard advertisement. Testing Assumption using Levene’s Test à Homogeneity of Variance. Test of Homogeneity of Variance Levene Statistic

$ spent during promotional period

df1

df2

Sig.

Based on Mean

1.190

1

498

.276

Based on Median

1.133

1

498

.288

Based on Median and with

1.133

1

497.000

.288

1.161

1

498

.282

adjusted df Based on trimmed mean

Levene’s Test Reporting: For the customers who spent money when there is a new promotion, compared to when there were standard ads, there was a difference. Spent money for new promotions, F (1, 498) = 1.19, but for unspent money for standard ads, F (1, 497) = 1.13, p < 0.01 (p. 152). 2. Required to test the two hypotheses: H1: the use of the medical treatment will lead to a significant reduction in the level of triglycerides of an individual; H2: the use of the medical treatment will lead to a significant reduction in the weight of an individual. For each of the hypotheses, which test will you use to test this hypothesis? Why is this test relevant? Test and report whether the assumptions that underlie this test are satisfied for each outcome that you are testing. Report the results using statistical terminology and interpret them. I will be doing a Dependent Means T-Test for this scenario. “This test is used when there are two experimental conditions and the same participants took part in both conditions of the experiment” (p. 325). In this scenario I must identify that the medical treatment led to a significant reduction in the level of triglycerides and weight of an individual. As you can see this is the same group that will be comparing their levels of triglycerides and weight over a period of time. Below are the results I found from testing the Dependent Means T-Test.


Paired Samples Statistics Mean Pair 1

Pair 2

N

Std. Deviation

Std. Error Mean

Triglyceride

138.44

16

29.040

7.260

Final triglyceride

124.38

16

29.412

7.353

Weight

198.38

16

33.472

8.368

Final weight

190.31

16

33.508

8.377

*It is important to note the Mean and Std. Error Mean column in this table. Paired Samples Correlations N Pair 1 Pair 2

Triglyceride & Final

Correlation

Sig.

16

-.286

.283

16

.996

.000

triglyceride Weight & Final weight

Paired Samples Test Paired Differences Mean

Std. Deviation

Std. Error Mean

t

df

Sig. (2-tailed)

95% Confidence Interval of the Difference Lower

Pair 1 Pair 2

Triglyceride - Final

Upper

14.063

46.875

11.719

-10.915

39.040

1.200

15

.249

8.063

2.886

.722

6.525

9.600

11.175

15

.000

triglyceride Weight - Final weight

*It is important to note the df and t column in this table. On average, individuals experienced significantly lower triglyceride levels in their final triglyceride test (M = 124.38, SE = 7.35) compared to the first initial triglyceride test (M = 138.44, SE = 7.26), t (15) = 1.2, p < 0.05, r = 0.29 (p. 333) On average, individuals experienced significantly lower weight in their final weight test (M = 190.31, SE = 8.38) compared to the first initial weight test (M = 196.38, SE = 8.37), t (15) = 11.18, p < 0.05, r = 0.94 (p. 333)


The hypotheses were supported by the data, which shows that the medical treatment did lead to lower levels of triglycerides and weight loss. It is important to note that there was more of a reduction in weight loss compared to triglyceride levels, which indicates that there should be an improvement for the triglyceride medical test so that they can both be effective. Testing Assumption using Tests of Normality à Kolmogorov-Smirnov and Shapiro-Wilk. Tests of Normality Kolmogorov-Smirnov Statistic Triglyceride Final triglyceride Weight Final weight

.194 .162 .156 .135

df

a

Shapiro-Wilk

Sig.

Statistic

df

Sig.

16

.111

.933

16

.277

16

.200

*

.941

16

.359

.200

*

.938

16

.320

.200

*

.949

16

.480

16 16

*. This is a lower bound of the true significance. a. Lilliefors Significance Correction

K-S Normality Reporting: The initial test for triglyceride, D(16) = 0.19, p < 0.05, and the final test for triglycerides, D(16) = 0.16, p < 0.001, were both a normal distribution (p. 148). In other words, there is no difference and the assumption of normality has been met. K-S Normality Reporting: The initial test for weight, D(16) = .156, p < 0.05, and the final test for weight, D(16) = 0.14, p < 0.001, were both significantly normal distribution (p. 148). In other words, there is no difference and the assumption of normality has been met.    


Assignment #5