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Data from a sample of 576 people in a city was used to estimate the mean monthly bill
paid by city residents for internet service to be $45 with a standard deviation of $0.48.
Enter a number in the box to complete the sentence below.
S
Use the formula M=1.96 x to calculate the margin of error for a 95% confidence
VN
interval, to the nearest hundredth, $
dollars.

Sagot :

Using the z-distribution, it is found that the margin of error for a 95% confidence interval is of $0.04.

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 sample.

The margin of error is given by:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In this problem, the values of the parameters are:

[tex]z = 1.96, \sigma = 0.48, n = 576[/tex].

Hence, the margin of error, in dollars, is given by:

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]M = 1.96\frac{0.48}{\sqrt{576}}[/tex]

M = 0.04.

More can be learned about the z-distribution at https://brainly.com/question/25890103

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