Question

You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places.

Answer 1

You intend to estimate a population mean μ from the following sample. 26.2 27.7 8.6 3.8 11.6 You believe the population is normally distributed. Find the 80% confidence interval.  Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places.

Answer:

The  Confidence interval = (8.98 , 22.18)

Step-by-step explanation:

From the given information:

mean = $\dfrac{ 26.2+ 27.7+ 8.6+ 3.8 +11.6 }{5}$

mean = 15.58

the standard deviation $\sigma$ = $\sqrt{\dfrac{\sum(x_i - \mu)^2 }{n}}$

the standard deviation = $\sqrt{\dfrac{(26.2 - 15.58)^2 +(27.7 - 15.58)^2 +(8.6 - 15.58)^2 + (3.8 - 15.58)^2 + (11.6 - 15.58)^2 }{5 } }$

standard deviation = 9.62297

Degrees of freedom df = n-1

Degrees of freedom df = 5 - 1

Degrees of freedom df = 4

For df  at 4 and 80% confidence level, the critical value t from t table  = 1.533

The Margin of Error M.O.E = $t \times \dfrac{\sigma}{\sqrt{n}}$

The Margin of Error M.O.E = $1.533 \times \dfrac{9.62297}{\sqrt{5}}$

The Margin of Error M.O.E = $1.533 \times 4.3035$

The Margin of Error M.O.E = 6.60

The  Confidence interval = ( $\mu \pm M.O.E$ )

The  Confidence interval = ( $\mu + M.O.E$ , $\mu - M.O.E$ )

The  Confidence interval = ( 15.58 - 6.60 , 15.58 + 6.60)

The  Confidence interval = (8.98 , 22.18)