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Sagot :
Answer: It's in the screenshot
Step-by-step explanation:
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
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