Respuesta :
Using the z-distribution, the 99% confidence interval for the true population mean textbook weight is (72.37, 81.63).
What is a z-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- z is the critical value.
- n is the sample size.
- [tex]\sigma[/tex] is the standard deviation for the population.
In this problem, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
The other parameters are given as follows:
[tex]\overline{x} = 77, \sigma = 12.2, n = 46[/tex]
Hence the lower and upper bound of the interval are given, respectively, by:
- [tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 77 - 2.575\frac{12.2}{\sqrt{46}} = 72.37[/tex]
- [tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 77 + 2.575\frac{12.2}{\sqrt{46}} = 81.63[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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