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.

Accepted Solution

Answer:   (1.1838, 1.1962)Step-by-step explanation:The formula we use to calculate the confidence interval for population mean ( if population standard deviation is not given)is given by :-[tex]\overline{x}\pm t*\dfrac{s}{\sqrt{n}}[/tex], where n= sample size s= sample standard deviation.[tex]\overline{x}[/tex]= sample meant* = Two-tailed critical t-value.Given : n= 25Degree of freedom : df = n-1 =24Significance level [tex]=\alpha=1-0.95=0.95[/tex]Now from students' t-distribution table , check the t-value for significance level [tex]\alpha/2=0.025[/tex] and df=24:t*=2.0639[tex]\overline{x}=1.19[/tex]  fluid ouncess= 0.015  fluid ouncesWe assume that the population is approximately normally distributedNow, the 95% two-sided confidence interval on the mean volume of syrup dispensed :-[tex]1.19\pm (2.0639)\dfrac{0.015}{\sqrt{25}}\\\\=1.19\pm (2.0639)(\dfrac{0.015}{5})\\\\=1.19\pm0.007728=(1.19-0.0061917,\ 1.19+0.0061917)\\\\=(1.1838083,\ 1.1961917)\approx(1.1838,\ 1.1962)[/tex]∴ The required confidence interval = (1.1838, 1.1962)