Question

question 9

Question 9 View Policies Current Attempt in Progress A random sample of n = 16 mid-sized cars tested for fuel consumption gave a mean of x = 28.90 miles per gallon with a standard deviation of s = 2.11 miles per gallon. Assuming that the miles per gallon given by all mid-sized cars have a normal distribution, find a 99% confidence interval for the population mean, . Round your answers to two decimal places. SHS i

Answer 1

Solution :

Given that,

$\bar x$ = 28.90

s = 2.11

n = 16

Degrees of freedom = df = n - 1 = 16 - 1 = 15

At 99% confidence level the t is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

t$\alpha$ /2,df = t0.005,15 =2.947

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.947 * (2.11 / $\sqrt$ 16)

= 1.10

Margin of error = 1.10

The 99% confidence interval estimate of the population mean is,

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

28.90 - 1.10 < $\mu$ < 28.90 + 1.10

27.80 < $\mu$ < 30.00

(27.80, 30.00 )