A manufacturer claims that the mean amount of juice in its 16 ounce bottles is 16.2 ounces. A consumer advocacy group wants to determine whether the mean amount is actually less than this. The mean volume of juice for a random sample of 70 bottles was 16.04 ounces with a standard deviation of 0.9 ounces. Find a 95% confidence interval to estimate the true mean amount of juice in these bottles.
A) What is the sample mean?
B) Standard deviation?
C) How many standard deviations do we need to capture 95% estimate?
D) Margin of error?
A, B) random sample of 70 bottles was 16.04 ounces with a standard deviation of 0.9 ounces.
C) ± 2 SD
D) Z = (score-mean)/SEm
SEm = SD/√n
A) The sample mean is the average volume of juice in the sample of 70 bottles, which is given as 16.04 ounces.
B) The standard deviation is a measure of the variability in the sample and is provided as 0.9 ounces.
C) To capture a 95% confidence estimate, we need to determine the number of standard deviations required. For a normal distribution, we can use the Z-score associated with a 95% confidence level, which is approximately 1.96. This means that we need to capture the sample mean within 1.96 standard deviations to have a 95% confidence.
D) The margin of error represents the range around the sample mean that contains the estimated true mean with a certain level of confidence. To calculate the margin of error, we can multiply the standard deviation by the Z-score. In this case, the margin of error can be calculated as:
Margin of error = Z-score * (Standard deviation / sqrt(sample size))
For a 95% confidence level (Z-score of 1.96) and a sample size of 70, the margin of error is:
Margin of error = 1.96 * (0.9 / sqrt(70))
Calculating this value gives us the margin of error.