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Leya [2.2K]
3 years ago
10

​Claim: High school teachers have incomes with a standard deviation that is more than ​$17, 500. A recent study of 136 high scho

ol teacher incomes showed a standard deviation of ​$18, 500.
A. Express the original claim in symbolic form.
B. Identify the null and the alternative hypotheses that should be used to arrive at the conclusion that supports the claim.
Mathematics
1 answer:
Scilla [17]3 years ago
4 0

Answer:

Part a

For this case the claim is:

\sigma >17500

And that represent the alternative hypothesis.

Part b: Null and alternative hypothesis

On this case we want to check if the population deviation is higher than 17500, so the system of hypothesis would be:

Null Hypothesis: \sigma \leq 17500

Alternative hypothesis: \sigma >17500

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{136-1}{17500^2} 18500^2 =150.869

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=136 represent the sample size

\alpha represent the confidence level  

s^2 =18500^2 represent the sample variance obtained

\sigma^2_0 =17500^2 represent the value that we want to test

Part a

For this case the claim is:

\sigma >17500

And that represent the alternative hypothesis.

Part b: Null and alternative hypothesis

On this case we want to check if the population deviation is higher than 17500, so the system of hypothesis would be:

Null Hypothesis: \sigma \leq 17500

Alternative hypothesis: \sigma >17500

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{136-1}{17500^2} 18500^2 =150.869

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