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Given the following hypotheses:Hou = 590590A random sample of 15 observations is selected from a normal population. The sample mean was 595 and the sample standarddeviation 8. Using the 0.05 significance level:a. State the decision rule. (Negative amount should be Indicated by a minus sign. Round your answers to three decimal places.)

User Shprink
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Answer

Explanation

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, since the null hypothesis, H₀, is already given as

H₀: μ = 590

In words, the null hypothesis would be that there is evidence from this sample that the population mean isn't significantly different from 590.

Then, the alternative hypothesis would be that the mean is significantly different from 590. That is,

Hₐ: μ ≠ 590

To know the decision rule, we need to first perform an hypothesis test on this data given

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic, which is given as

t = (x - μ)/σₓ

x = sample mean = 595

μ = The standard we are comparing against = 590

σₓ = standard error of the mean = (σ/√n)

σ = sample standard deviation = 8

where n = Sample size = 15

σₓ = (8/√15) = 2.066

t = (x - μ)/σₓ

t = (595 - 590)/2.066

t =

User Michael Roland
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