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The average (mean) annual income was less than $50,000
Solution:
Step1: State the Null and Alternate Hypothesis:
Null Hypothesis: The average (mean) annual income was greater than or equal to $50,000
H_0: μ≥50000
Alternate Hypothesis: The average (mean) annual income was less than $50,000.
H_a: μ 30 we shall use z-test for mean to test the given hypothesis.
As the alternative hypothesis is Ha:μ0.40 , the given test is a one-tailed (upper-tailed) z-test.

Step3: Critical Value and Decision Rule:
The critical value for significance level, α=0.05 for an upper-tailed z-test is given as 1.645.
Decision Rule: Reject H_0,if z-statistic>1.645

Step4: Test Statistic (MINITAB OUTPUT):
Number of Events (Urban Customers) = 21
Number of Trials (Customers) = 50
Test and CI for One Proportion

Test of p = 0.4 vs p > 0.4

95% Lower
Sample X N Sample p Bound Z-Value P-Value
1 21 50 0.420000 0.305190 0.29 0.386

Using the normal approximation.

Step5: Interpretation of Results and Conclusion:
Since the P-value (0.386) is greater than the significance level (0.05), we fail to reject the null hypothesis. The p-value implies the probability of rejecting a true null hypothesis.
Thus, at a significance level of 0.05, there is no sufficient evidence to support the claim that the true population proportion of customers who live in an urban area is greater than 40%.

Confidence Interval (MINITAB OUTPUT):
Test and CI for One Proportion

Sample X N Sample p 95% CI
1 21 50 0.420000 (0.283195, 0.556805)

The 95% lower confidence limit is 0.28. Since 0.42 is greater than the 95% lower confidence limit, hence we cannot support the claim that the true population proportion of customers who live in an urban area is greater than 40%.