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Jeremy is conducting a survey about his coworkers’ in-office water consumption to encourage management to install more water dispensers at their location. He found that the population mean is 112.5 ounces with a standard deviation of 37.5. Jeremy has a sample size of 96. Complete the equation that Jeremy can use to find the interval in which he can be 99.7% sure that the sample mean will lie. 37.5 112.5 75 96 150 9.8

Respuesta :

Using the z-distribution, as we have the standard deviation for the population, it is found that the 99.7% confidence interval is given by:

[tex]112.5 \pm 3\frac{37.5}{\sqrt{96}}[/tex]

What is a t-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.
  • t is the critical value.
  • n is the sample size.
  • [tex]\sigma[/tex] is the standard deviation for the population.

99.7% confidence level, hence[tex]\alpha = 0.997[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.997}{2} = 0.9985[/tex], so [tex]z = 3[/tex].

The other parameters are:

[tex]\mu = 112.5, \sigma = 37.5, n = 96[/tex]

Hence, the interval is:

[tex]112.5 \pm 3\frac{37.5}{\sqrt{96}}[/tex]

To learn more about the z-distribution, you can check https://brainly.com/question/25890103

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