Suppose a researcher is interested in understanding the variation in the price of store brand milk.A random sample of 36 grocery stores selected from a population and the mean price of store brand milk is calculated.The sample mean is $3.13 with a standard deviation of $0.23.Construct a 95% confidence interval to estimate the population mean.
A) $3.03 to $3.23
B) $3.12 to $3.14
C) $3.05 to $3.21
D) $2.90 to $3.36
E) $3.06 to $3.20

Question

Quizplus · Accepted Answer

We can use the formula: 
Confidence interval= sample mean ± (critical value) x (standard error)
where critical value is obtained from t-distribution table at (n-1) degrees of freedom and desired confidence level, standard error= standard deviation/√n.

Here, n=36, sample mean= $3.13, standard deviation= $0.23.
Degrees of freedom= (n-1) = 35
From the t-distribution table at 35 degrees of freedom and 95% confidence level, the critical value is 2.03 (rounded off).
Standard error= $0.23/√36= $0.0383 (rounded off)

Substituting the values, we get:
Confidence interval= $3.13 ± 2.03 x $0.0383
Confidence interval= $3.13 ± $0.0778
Confidence interval= ($3.05, $3.21)

Thus, the 95% confidence interval for the population mean price of store brand milk is from $3.05 to $3.21.

Therefore, the best choice is C.

Suppose a Researcher Is Interested in Understanding the Variation in the Price