Question

Kolbe, product manager for a line of shoes, is wondering whether to introduce his product line into a new market area. A recent survey of a random sample of 500 households in that market showed a mean household income of $34,000 with a standard deviation of $2,000. On the basis of past experience and of comprehensive studies in current market areas, Kolbe believes the product line will be profitable only in markets where the mean household income (across all households) is greater than $30,000. Should Kolbe introduce the product line into the new market?

Answer 1

Since n is very large, we use the Z test

1 Tailed Z test, Single Mean

Given: = $30,000, = $34,000, s = $2,000, n = 500, = 0.05 (Default)

The Hypothesis:

H0: = 30,000: The mean household income is equal to 30,000 dollars.

Ha: > 30,000: The mean household income isgreater than 30,000 dollars.

This is a 1 tailed test (Right tailed)

The Test Statistic: The test statistic is given by the equation:

Z observed = 44.72

The p Value: The p value for Z = 44.72, p value = 0.000

The Critical Value: The critical value (1 Tail) at = 0.05, tcritical= +1.645

The Decision Rule:   If Zobserved is > Zcritical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision: Since Zobserved (44.72) is greater than Z critical, We Reject H0.

Also since P value (0.000) is < (0.05) , We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the mean household income is equal to 30,000 dollars.Kolbe should introduce the product line into the new market.