Question

A bottling plant produces 1 liter bottles of soda. The actual distribution of volumes of soda dispensed to bottles is Normal, with mean μ and standard deviation σ = 0.05 liter. We randomly select 6 bottles andmeasure the volume of soda in each. The results of these 6 measurements (all in liter units) are 1.05 1.04 1.01 1.06 0.94 0.99. Based on these data, a 90% confidence interval for μ is

Answer 1

 μ = population mean
σ = 0.05, the population standard deviation

Sample size, n = 6
Sample mean, $\bar{x}$ = (1.05+1.04+1.01+1.06+0.94+0.99)/6 = 1.015

Because the sample size is less than 30, the confidence interval is
$\bar{x} \pm t^{*} \frac{\sigma}{ \sqrt{n} }$
where
t* = 2.015, fromm the t-distribution with dof = 6-1 = 5.

The confidence interval for μ is
1.015 +/- 2.015(0.05/√6) = 1.015 +/- 0.0411 = (0.974, 1.056)

Answer: (0.974, 1.056) liters

Answer 2

Answer:

μ = population mean

σ = 0.05, the population standard deviation

Sample size, n = 6

Sample mean, \bar{x}

x

ˉ

= (1.05+1.04+1.01+1.06+0.94+0.99)/6 = 1.015

Because the sample size is less than 30, the confidence interval is

\bar{x} \pm t^{*} \frac{\sigma}{ \sqrt{n} }

x

ˉ

±t

∗

n

σ

where

t* = 2.015, fromm the t-distribution with dof = 6-1 = 5.

The confidence interval for μ is

1.015 +/- 2.015(0.05/√6) = 1.015 +/- 0.0411 = (0.974, 1.056)

Answer: (0.974, 1.056) liters