In speaking with my management team they have speculated that for the data set the average annual income for the customers selected is less than $50,000, it is also speculated that the true population proportion of customers who live in the urban area exceeds 40%, That the average number of years our customers live in their current home is less than 13 years, and that the average credit balance for our suburban customers is more than $4,300.
Testing the hypothesis that the mean annual income was less than $50,000
One-Sample Z: Income ($1000)

Test of mu = 50 vs < 50
The assumed standard deviation = 14.64 95% Upper
Variable N Mean StDev SE Mean Bound Z P
Income ($1000) 50 43.74 14.64 2.07 47.15 -3.02 0.001
Test of mu = 50 vs not = 50
The assumed standard deviation = 14.64

Variable N Mean StDev SE Mean 95% CI Z P
Income ($1000) 50 43.74 14.64 2.07 (39.68, 47.80) -3.02 0.002

The manager speculated the average income is less than $50,000, according to my stat P value is lower than alpha(.05) this means that the analysis the manager's claim is supported with a 95% confidence. Meaning that the null hypothesis of average is greater or equal to $50,000 is rejected and the alternative hypothesis is supported. The confidence interval at a 95% confidence is between 39.68,47.80.
Testing the hypothesis that the true population proportion of customers who live in an urban area exceeds 40%
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

Test and CI for One Proportion

The manager speculated the true population of customers who live in an urban area exceeds 40%,