Question

Construct the indicated confidence interval for the population mean

muμ using the​ t-distribution. Assume the population is normally distributed.

cequals=0.99 x overbarxequals=13.6​, sequals=0.56 nequals=15

Round to one decimal place as needed

Answer 1

Given that,

$\bar x$ = 13.6

s =0.56

n = 15

Degrees of freedom = df = n - 1 = 15- 1 = 14

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} = t_0.005,14 = 2.977

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

= 2.977* (0.56 / $\sqrt$ 15) = 0.4

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

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

13.6 - 0.4 < $\mu$ < 13.6+ 0.4

13.2 < $\mu$ < 14.0

(13.2 ,14.0 )