A quality control expert at a drink bottling factory took a random sample of 9 bottles from a
production run of over 1,000 bottles and measured the amount of liquid in each bottle in the
sample. The sample data were roughly symmetric with a mean of 505 mL and a standard
deviation of 6 mL.
Based on this sample, which of the following is a 95% confidence interval for the mean
amount of liquid per bottle (in mL) in this production run?

Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval for the mean amount of liquid per bottle (in mL) in this production run is (500.39, 509.61).

What is a t-distribution confidence interval?

The confidence interval is:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

In which:

[tex]\overline{x}[/tex] is the sample mean.

t is the critical value.

n is the sample size.

s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 9 - 1 = 8 df, is t = 2.306.

The other parameters are given as follows:

[tex]\overline{x} = 505, s = 6, n = 9[/tex]

Hence:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 505 - 2.306\frac{6}{\sqrt{9}} = 500.39[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 505 + 2.306\frac{6}{\sqrt{9}} = 509.61[/tex]

The 95% confidence interval for the mean amount of liquid per bottle (in mL) in this production run is (500.39, 509.61).

More can be learned about the t-distribution at https://brainly.com/question/16162795