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A study of a local high school tried to determine the mean amount of money that each
student had saved. The study surveyed a random sample of 69 students in the high
school and found a mean savings of 5000 dollars with a standard deviation of 1400
dollars. At the 95% confidence level, find the margin of error for the mean, rounding
to the nearest whole number. (Do not write ).

A Study Of A Local High School Tried To Determine The Mean Amount Of Money That Each Student Had Saved The Study Surveyed A Random Sample Of 69 Students In The class=

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Using the t-distribution, it is found that the margin of error for the 95% confidence interval is of $336.

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.
  • t is the critical value.
  • n is the sample size.
  • s is the standard deviation for the sample.

The margin of error is given by:

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

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 69 - 1 = 68 df, is t = 1.9955.

The standard deviation and sample size are given, respectively, by:

s = 1400, n = 69.

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

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

[tex]M = 1.9955\frac{1400}{\sqrt{69}}[/tex]

M = 336.

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

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