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 27 beverages was found to have a mean syrup content 1.20 fluid ounce and a standard deviation of s = 0.015 fluid ounce. Find a 95% CI on the mean volume of syrup dispensed.t0.025,26=2.056𝑡0.025,26=2.056, t0.025,27=2.052𝑡0.025,27=2.052, t0.05,26=1.706𝑡0.05,26=1.706a.[1.194, 1.206]b.[1.094 , 1.106]c.[1.199, ∞∞]
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 27 beverages was found to have a mean syrup content 1.20 fluid ounce and a standard deviation of s = 0.015 fluid ounce. Find a 95% CI on the mean volume of syrup dispensed.t0.025,26=2.056𝑡0.025,26=2.056, t0.025,27=2.052𝑡0.025,27=2.052, t0.05,26=1.706𝑡0.05,26=1.706a.[1.194, 1.206]b.[1.094 , 1.106]c.[1.199, ∞∞]
Solution
To find a 95% confidence interval (CI) for the mean volume of syrup dispensed, we can use the formula for a confidence interval:
CI = x̄ ± t(s/√n)
where: x̄ is the sample mean, t is the t-value from the t-distribution table (which depends on the desired level of confidence and the degrees of freedom, which is n-1), s is the sample standard deviation, and n is the sample size.
Given in the problem, we have: x̄ = 1.20 fluid ounces, s = 0.015 fluid ounces, n = 27.
The t-value for a 95% confidence level and 26 degrees of freedom (n-1 = 27-1 = 26) is approximately 2.056 (given in the problem).
Substituting these values into the formula, we get:
CI = 1.20 ± 2.056(0.015/√27)
Calculating the expression inside the parentheses gives approximately 0.0057.
So, the 95% confidence interval is 1.20 ± 0.0057, or [1.1943, 1.2057].
Rounding to three decimal places, the 95% confidence interval is [1.194, 1.206]. So, the correct answer is (a) [1.194, 1.206].
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