Question

an economics professor randomly selected 100 millionaires in the united states. the average age of these millionaires was 54.8 years with a standard deviation of 7.9 years. what is the 95% confidence interval for the mean age, of all united states millionaires. A.) A confidence interval for the mean using a table value from the standard normal distribution B.) A confidence interval for the mean using a table value from the t distribution C.) A confidence interval for a proportion using a table value from the standard normal distribution D.) A confidence interval for proportion using a table value from the t distribution E.) None of the above

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 54.8

Population standard deviation =   $\sigma$ = 7.9

Sample size = n =100

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.96   ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

Margin of error = E = Z$\alpha$/2 * ( $\sigma$ /$\sqrt$n)

= 1.96 * (7.9 /  $\sqrt$100 )

E= 1.5484
At 95% confidence interval estimate of the population mean
is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

54.8 - 1.5484 <  $\mu$ < 54.8 + 1.5484

53.2516<  $\mu$ < 56.3484

( 53.2516, 56.3484 )

correct option A.) A confidence interval for the mean using a table value from the standard normal distribution