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An admission officer of an educated institution wants to know the mean age of all entering Mathematics Majors. He computed a mean age of 18 years and a standard deviation of 1.2 years on a random sample of 25 entering math majos, coming from a normally distributed population. With 99% confidence, find the point estimate and the interval estimate of the population mean.

Respuesta :

Using the t-distribution, as we have the standard deviation for the sample, it is found that:

  • The point estimate of the population mean is of 18 years.
  • The 99% interval estimate of the population mean is (17.33, 18.67).

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, and is also the point estimate of the population 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 99% confidence interval, with 25 - 1 = 24 df, is t = 2.797.

The other parameters are given as follows:

[tex]\overline{x} = 18, s = 1.2, n = 25[/tex].

Hence, the bounds of the interval are given by:

[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 18 - 2.797\frac{1.2}{\sqrt{25}} = 17.33[/tex]

[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 18 + 2.797\frac{1.2}{\sqrt{25}} = 18.67[/tex]

The 99% interval estimate of the population mean is (17.33, 18.67).

More can be learned about the t-distribution at https://brainly.com/question/16162795

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