Question

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 ).

Answer 1

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:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

• $\overline{x}$ 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:

$M = t\frac{s}{\sqrt{n}}$

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:

$M = t\frac{s}{\sqrt{n}}$

$M = 1.9955\frac{1400}{\sqrt{69}}$

M = 336.

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

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