Data on salaries in the public school system are published annually by a teacher

Question

Data on salaries in the public school system are published annually by a teachers' association. The mean annual salary of (public) classroom teachers is $51.3 thousand. Assume a standard deviation of $8.0 thousand. Complete parts (a) through (e) below.

d. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1000?

e. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 256 classroom teachers will be at most $1000?

Explanation / Answer

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Data on salaries in the public school system are published annually by ateachers’ association. The mean annual salary of (public) classroom teachers is $51.3 thousand. Assume a standard deviation of $8.0 thousand. Complete parts (a) through (e) below.

d. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1000?

Solution:

Here, we are given the sampling error = $1000 or $1 thousand

Please note that the mean and standard deviation are given in ‘thousands’.

This means, (xbar – population mean) = 1

The formula for z score is given as below:

Z = (xbar – population mean) / [SD/sqrt(n)]

Where, n is the sample size

Z = 1 / [8/sqrt(64)] = 1/[8/8] = 1/1 = 1

P(Z<1) = 0.841345

Required probability = 0.841345

e. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 256 classroom teachers will be at most $1000?

Solution:

Here, we are given the sampling error = $1000 or $1 thousand

Please note that the mean and standard deviation are given in ‘thousands’.

This means, (xbar – population mean) = 1

The formula for z score is given as below:

Z = (xbar – population mean) / [SD/sqrt(n)]

Where, n is the sample size

Z = 1 / [8/sqrt(256)] = 1/[8/16] = 1/(1/2) = 2

P(Z<1) = 0.97725

Required probability = 0.97725

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