MATH SOLVE

4 months ago

Q:
# 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

Accepted Solution

A:

μ = population mean

σ = 0.05, the population standard deviation

Sample size, n = 6

Sample mean, [tex]\bar{x}[/tex] = (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

[tex]\bar{x} \pm t^{*} \frac{\sigma}{ \sqrt{n} } [/tex]

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

σ = 0.05, the population standard deviation

Sample size, n = 6

Sample mean, [tex]\bar{x}[/tex] = (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

[tex]\bar{x} \pm t^{*} \frac{\sigma}{ \sqrt{n} } [/tex]

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