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Regular consumption of presweetened cereals contributes to tooth decay, heart disease, and other degenerative diseases, according to studies conducted by Dr. W. H. Bowen of the National Institute of Health and Dr. J. Yudben, Professor of Nutrition and Dietetics at the University of London. In a random sample consisting of 20 similar single servings of Alpha-Bits, the average sugar content was 11.3 grams with a standard deviation of 2.45 grams. Assuming that the sugar contents are normally distributed, construct a 95% confidence interval for the mean sugar content for single servings of Alpha-Bits.

Respuesta :

Using the t-distribution, the 95% confidence interval for the mean sugar content for single servings of Alpha-Bits is (10.15, 12.45).

In this problem, we have that:

  • Sample of 20, thus [tex]n = 20[/tex].
  • Sample mean of 11.3 grams, thus [tex]\overline{x} = 11.3[/tex].
  • Sample standard deviation of 2.45 grams, thus [tex]s = 2.45[/tex].

The confidence interval is:

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

First, we find the number of degrees of freedom, which is one less than the sample size, thus df = 19.

Now, we have to find the critical value for a 95% confidence interval with 19 df, thus, looking at the t-table or using a calculator, t = 2.093.

Then:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 11.3 - 2.093\frac{2.45}{\sqrt{20}} = 10.15[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 11.3 + 2.093\frac{2.45}{\sqrt{20}} = 12.45[/tex]

The interval is (10.15, 12.45).

A similar problem is given at https://brainly.com/question/24232455

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