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An SRS of ten brands of breakfast cereals is tested for the number of calories per serving. The following data result: 185, 190, 195, 200, 205, 205, 210, 210, 225, 230. Find a 95% confidence interval estimate for the mean number of calories for servings of breakfast cereals.

Sagot :

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

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