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Question 3 0/3 pts 399 Details You wish to test the following claim (H) at a significance level of a 0.002 H: = 55.3 H: 55.3
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Answer #1

Given Data

Sample mean is X = 56

Sample standard deviation is s = 7.5

Sample size = n = 28

significance level is = α = 0.002

Null and alternative hypotheses need to be tested

Ho: μ = 55.3

Ha: μ ≠ 55.3

This corresponds to a two-tailed test

At the significance level is α=0.002, and the critical value for a two-tailed test is

tc​ (lower)= - 3.421 using excel =T.INV(0.001,27)

tc​ (upper)= 3.421 using excel =T.INV(0.999,27)

Rejection region will be t<-3.421 and t>3.421

Test Statistics

t = \frac{X-\mu}{\frac{s}{\sqrt{n}}}

t = \frac{56.9-55.3}{\frac{7.5}{\sqrt{28}}} = 1.129

(1) Test Statistics = t = 1.129

(2) p value is calculated as p = 0.2688, using excel =T.DIST.2T(1.129,27)

(3) p = 0.2688 > α=0.002,, This mean p value is greater than α.

(4) Since t test value is greater than tc lower and less than tc upper

-3.421<t = 1.129<3.421

So null hypothesis is not rejected

(5) As such Final conclusion is that There is sufficient evidence to warrant rejection of the claim that population mean is not equal to 55.3, So 1st option is correct

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