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Q3. Hypothesis: Informing people about recycling causes them to recycle more. Study design: 50 households were...

Q3. Hypothesis: Informing people about recycling causes them to recycle more.

Study design: 50 households were randomly assigned to a treatment group where they were

informed by letter about proper recycling habits and its benefit on environment, while 50

different households were randomly assigned to a control group that did not receive such a letter.

After 3 months, the weekly average recycling amount in the treatment group was 12.4 lbs (

sd

=

2.5), while the weekly average recycling amount in the control group was 3.7 lbs (

sd

= 1.1)

d. Determine the appropriate test: z-test or t-test. Explain why you chose that test.

e. Calculate the appropriate test statistic

f. Decide whether you should reject the null hypothesis.

Solutions

Expert Solution

Q3. Hypothesis: Informing people about recycling causes them to recycle more.

Study design: 50 households were randomly assigned to a treatment group where they were

informed by letter about proper recycling habits and its benefit on environment, while 50

different households were randomly assigned to a control group that did not receive such a letter.

After 3 months, the weekly average recycling amount in the treatment group was 12.4 lbs (sd=2.5), while the weekly average recycling amount in the control group was 3.7 lbs (sd= 1.1)

d. Determine the appropriate test: z-test or t-test. Explain why you chose that test.

T test is used because population standard deviation is not known.

Two sample t test

Ho: µ1 = µ2   H1: µ1 > µ2

Upper tail t test

e. Calculate the appropriate test statistic

 

t=22.5234

f. Decide whether you should reject the null hypothesis.

DF = n1+n2-2 =98

Table value of t with 98 DF at 0.05 level = 1.6606

Rejection Region: Reject Ho if or t > 1.6606

Calculated t = 22.5234 falls in the rejection region

The null hypothesis is rejected.

We conclude that informing people about recycling causes them to recycle more.

Pooled-Variance t Test for the Difference Between Two Means

(assumes equal population variances)

Data

Hypothesized Difference

0

Level of Significance

0.05

Population 1 Sample

Sample Size

50

Sample Mean

12.4

Sample Standard Deviation

2.5

Population 2 Sample

Sample Size

50

Sample Mean

3.7

Sample Standard Deviation

1.1

Intermediate Calculations

Population 1 Sample Degrees of Freedom

49

Population 2 Sample Degrees of Freedom

49

Total Degrees of Freedom

98

Pooled Variance

3.7300

Standard Error

0.3863

Difference in Sample Means

8.7000

t Test Statistic

22.5234

Upper-Tail Test

Upper Critical Value

1.6606

p-Value

0.0000

Reject the null hypothesis


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