Question

Case 1: The company XYZ, is dedicated to the production of coke, a campaign to reposition its products with the aim of increasing its presence in the segments with the highest socioeconomic level. The median monthly income before the campaign was known to be $ 8,000. The campaign consists of changing the label of the drinks, using other advertising channels and increasing the price of their products. At the end of the campaign, a random sample of 80 clients was requested, who had their average annual salary of $ 9,500 and a standard deviation of $ 1,800. With a significance level of 5%
Was the campaign successful? Was there a significant increase in the average monthly income of customers? Justify your answer.

Is this correct regarding the case? n = 80 clients       X ̂~N(μ=$9,500,σ=$1,800)       ∝=0.05

It should be with T because I think we don't know the Standard Deviation. What should be done?
The traditional method or an alternative method? What else should I do?

Answer 1

Yes it Must be done with the t statistic because population standard deviation is unknown.

X ̂~N(μ=$9,500,s=$1,800)   

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA:

LEVEL OF SIGNIFICANCE =0.05

test statistic t= xbar-mu/sigma/sqrt(n)

= 9500-8000/1800/sqrt(80)

= 1500/1800/8.94

= 1500/201.34

t=7.45

degrees of freedom= n-1=80-1=79

Pvalue= 0.0000

Since P value SMALLER THAN 0.05 THEREFORE SIGNIFICANT

DECISION: REJECT NULL HYPOTHESIS H0.

CONCLUSION: We have sufficient evidence to conclude that  there is a significant increase in the average monthly income of customers at 0.05 level of significance.