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3 years ago

6

Ramona is counting the posts between mile markers on the highway. In 1 mile, she counts 33 posts if the posts are evenly spaced,

how many feet are they apart? (1 mile = 5,280 feet)

1 answer:

Yuri [45]3 years ago

6 0

5280/33 = 160 ft apart

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Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).

<h3>What are the mean and the standard error for each sample?</h3>

Considering the data given:

$\mu_S = 3.8, n = 175, s_S = \frac{1.7}{\sqrt{175}} = 0.1285$

$\mu_N = 4.8, n = 152, s_N = \frac{1.2}{\sqrt{152}} = 0.0973$

<h3>What is the mean and the standard error for the distribution of differences?</h3>

The mean is the subtraction of the means, hence:

$\overline{x} = \mu_S - \mu_N = 3.8 - 4.8 = -1$

The standard error is the square root of the sum of the variances of each sample, hence:

$s = \sqrt{s_S^2 + s_N^2} = \sqrt{0.1285^2 + 0.0973^2} = 0.1612$

<h3>What is the confidence interval?</h3>

It is given by:

$\overline{x} \pm zs$

We have a 95% confidence interval, hence the critical value is of z = 1.96.

Then, the bounds of the interval are given as follows:

$\overline{x} - zs = -1 - 1.96(0.1612) = -1.3$

$\overline{x} + zs = -1 + 1.96(0.1612) = -0.7$

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

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