weekly average expenditure for 200 adults with a deviation of 14.50, confidence interval of 95%

Question

200 adults surveyed standard deviation of $14.50.  95% confidence interval and $35 weekly expenditure

Answer 1

Applying the confidence level of 95%, a=1-0.95=0.05 and the critical probability is 1-a/2=0.975. The number of degrees of freedom is 200-1=199, and the standard deviation (SD) is $14.50. The standard error (SE) is therefore 14.5/sqrt(200)=1.025. We can work out the critical value from t score tables: 1.96 given df>150 and the critical probability of 0.975. The margin of error is SE*1.96=2.01 approx., and the weekly average expenditure is $35+2.0 approx.