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A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. The 95% confidence interval for the true mean number of pushups that can be done is _____.

Respuesta :

Using the t-distribution, we have that the 95% confidence interval for the true mean number of pushups that can be done is (9, 21).

For this problem, we have the standard deviation for the sample, thus, the t-distribution is used.

  • The sample mean is of 15, thus [tex]\overline{x} = 15[/tex].
  • The sample standard deviation is of 9, thus [tex]s = 9[/tex].
  • The sample size is of 10, thus [tex]n = 10[/tex].

First, we find the number of degrees of freedom, which is the one less than the sample size, thus df = 9.

Then, looking at the t-table or using a calculator, we find the critical value for a 95% confidence interval, with 9 df, thus t = 2.2622.

The margin of error is of:

[tex]M = t\frac{s}{n}[/tex]

Then:

[tex]M = 2.2622\frac{9}{\sqrt{10}} = 6[/tex]

The confidence interval is:

[tex]\overline{x} \pm M[/tex]

Then

[tex]\overline{x} - M = 15 - 6 = 9[/tex]

[tex]\overline{x} + M = 15 + 6 = 21[/tex]

The confidence interval is (9, 21).

A similar problem is given at https://brainly.com/question/25157574

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