Using the t-distribution, as we have the standard deviation for the sample, it is found that the 95% confidence interval estimate for the mean number of calories for servings of breakfast cereals is (195.3, 215.7).
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.
From the sample and the significance level of 0.05, we have that the parameters are given by:
[tex]\overline{x} = 205.5, s = 14.2, n = 10, t = 2.2622[/tex]
Hence:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 205.5 - 2.2622\frac{14.2}{\sqrt{10}} = 195.3[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 205.5 + 2.2622\frac{14.2}{\sqrt{10}} = 215.7[/tex]
The 95% confidence interval estimate for the mean number of calories for servings of breakfast cereals is (195.3, 215.7).
More can be learned about the t-distribution at https://brainly.com/question/16162795
