1. Coco Martinez is the Head of the of the Prow, Vince, Sean & Noe Marketing, a

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1. Coco Martinez is the Head of the of the Prow, Vince, Sean & Noe Marketing, a leading marketing consulting fim which provides marketing and advertising consulting services. Coco is currently facing with four marketing-related problems which must be decided soon, but must be supported with factual evidences. When not given proper decision and interpretation, this would put his consultancy business into a dilemma of labeling "erratic" and "irresolute" by his clients. Help him solve several issues on marketing to win the trust of his clients, using the approprnate test of hypothesis between means. a Celebrity endorsement of products is a common technique. A group of 98 people was shown an ad containing a celebrity endorsement, and a second group of 98 was shown the same ad but using an unknown actor. Each participant rated the commercial's believability on a scale of 0 (not believable) to 10 (very believable) Results were as follows: Believability of Ad Type of Ad Celebrity ad Non-celebrity ad Mean 3.82 3.97 Standard Deviation 2.63 2.51 Is there sufficient evidence to indicate that the use of a celebrity endorsement results in a true mean believability rating that differs from the true mean rating for non- celebrity endorsements? Use a level 0.05 test bl Twenty-four infants paired according to birth weight were used to compare an enriched formula with a standard formula. Weight gains (in grams) are given. PairEnriched Formula Standard Formula 3,604 2,950 3,344 4,022 4,316 3,077 2,988 5,200 4,653 3,218 3,461 2,999 3,140 3,100 2,810 3,761 3,774 2,630 2,988 4,388 5,133 3,621 3,679 3,500 6 10 12 Test Ho: Hd0 against this alternative hypothesis at significance level. 05. Assume populations to be approximately distributed with equal variances

Explanation / Answer

Question 1

Part a

Here, we have to use independent samples t test for difference between population means. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: There is no significant difference in the mean believability rating if ad uses the celebrity and ad does not use the celebrity.

Alternative hypothesis: Ha: There is a significant difference in the mean believability rating if ad uses the celebrity and ad does not use the celebrity.

The level of significance or alpha value is given as 0.05.

Results for this test by using excel are summarised in the following table:

[Excel Note: Open Excel > Data > Data Analysis > t-test assuming equal population variances]

[If data is summarised, above excel data analysis tool would not be work. In this case, go to excel settings > get ad on > get PHStat or other ad on, this will give results for summarised data.]

[Excel data analysis tools are only used for un-summarised data.]

Pooled-Variance t Test for the Difference Between Two Means

(assumes equal population variances)

Data

Hypothesized Difference

Level of Significance

0.05

Population 1 Sample

Sample Size

Sample Mean

3.82

Sample Standard Deviation

2.63

Population 2 Sample

Sample Size

Sample Mean

3.97

Sample Standard Deviation

2.51

Intermediate Calculations

Population 1 Sample Degrees of Freedom

Population 2 Sample Degrees of Freedom

Total Degrees of Freedom

194

Pooled Variance

6.6085

Standard Error

0.3672

Difference in Sample Means

-0.1500

t Test Statistic

-0.4084

Two-Tail Test

Lower Critical Value

-1.9723

Upper Critical Value

1.9723

p-Value

0.6834

Do not reject the null hypothesis

We do not reject the null hypothesis that there is no significant difference in the mean believability rating if ad uses the celebrity and ad does not use the celebrity.

There is insufficient evidence to conclude that there is a significant difference in the mean believability rating if ad uses the celebrity and ad does not use the celebrity.

Part b

Here, we have to use paired t test for checking the given claim. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: Average weight gain in grams with a enriched formula is same as average weight gain in grams with a standard formula. There is no any significant difference in average weight gains with an enriched formula and standard formula.

Alternative hypothesis: Ha: Average weight gain in grams with an enriched formula is not same as average weight gain in grams with a standard formula. There is a significant difference in average weight gains with an enriched formula and standard formula.

H0: µd = 0 vs. Ha: µd ? 0 (Two tailed test)

? = 0.05

Excel output for this test is given as below:

Enriched formula

Standard formula

(Di - DBar)^2

3604

3140

464

126025

2950

3100

-150

67081

3344

2810

534

180625

4022

3761

261

23104

4316

3774

542

187489

3077

2630

447

114244

2988

11881

5200

4388

812

494209

4653

5133

-480

346921

3218

3621

-403

262144

3461

3679

-218

106929

2999

3500

-501

372100

Dbar =

109

t-Test: Paired Two Sample for Means

Enriched formula

Standard formula

Mean

3652.666667

3543.666667

Variance

544439.5152

491543.1515

Observations

Pearson Correlation

0.799850744

Hypothesized Mean Difference

t Stat

0.827056012

P(T<=t) one-tail

0.212891446

t Critical one-tail

1.795884814

P(T<=t) two-tail

0.425782891

t Critical two-tail

2.200985159

Paired t Test

Data

Hypothesized Mean Difference

Level of significance

0.05

Intermediate Calculations

Sample Size

DBar

109.0000

Degrees of Freedom

456.5435

Standard Error

131.7928

t Test Statistic

0.8271

Two-Tail Test

Lower Critical Value

-2.2010

Upper Critical Value

2.2010

p-Value

0.4258

Do not reject the null hypothesis

We do not reject the null hypothesis that Average weight gain in grams with a enriched formula is same as average weight gain in grams with a standard formula. There is no any significant difference in average weight gains with an enriched formula and standard formula.

Pooled-Variance t Test for the Difference Between Two Means

(assumes equal population variances)

Data

Hypothesized Difference

Level of Significance

0.05

Population 1 Sample

Sample Size

Sample Mean

3.82

Sample Standard Deviation

2.63

Population 2 Sample

Sample Size

Sample Mean

3.97

Sample Standard Deviation

2.51

Intermediate Calculations

Population 1 Sample Degrees of Freedom

Population 2 Sample Degrees of Freedom

Total Degrees of Freedom

194

Pooled Variance

6.6085

Standard Error

0.3672

Difference in Sample Means

-0.1500

t Test Statistic

-0.4084

Two-Tail Test

Lower Critical Value

-1.9723

Upper Critical Value

1.9723

p-Value

0.6834

Do not reject the null hypothesis

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