Question

A postmix beverage machine is adjusted to release a certain amount of syrup into a chamber where it is mixed with carbonated water. A random sample of 25 beverages was found to have a mean syrup content of x¯=1.19 fluid ounces and the sample standard deviation is s=0.015 fluid ounces. Find a 95% two-sided confidence interval on the mean volume of syrup dispensed. Assume population is approximately normally distributed.

Answer 1

Answer: (1.1838, 1.1962)

Explanation:

The formula we use to calculate the confidence interval for population mean ( if population standard deviation is not given)is given by :-

`\overline{x}\pm t*(s)/(√(n))`,

where n= sample size

s= sample standard deviation.

`\overline{x}`= sample mean

t* = Two-tailed critical t-value.

Given : n= 25

Degree of freedom : df = n-1 =24

Significance level
`=\alpha=1-0.95=0.95`

Now from students' t-distribution table , check the t-value for significance level
`\alpha/2=0.025` and df=24:

t*=2.0639

`\overline{x}=1.19` fluid ounces

s= 0.015 fluid ounces

We assume that the population is approximately normally distributed

Now, the 95% two-sided confidence interval on the mean volume of syrup dispensed :-

`1.19\pm (2.0639)(0.015)/(√(25))\\\\=1.19\pm (2.0639)((0.015)/(5))\\\\=1.19\pm0.007728=(1.19-0.0061917,\ 1.19+0.0061917)\\\\=(1.1838083,\ 1.1961917)\approx(1.1838,\ 1.1962)`

∴ The required confidence interval = (1.1838, 1.1962)