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n=90, Y =18.8, S =15.3, α =0.05⇒tn−1,α/2 = t89,.025 =1.987 H0 : µ =21.7 versus HA...

n=90, Y =18.8, S =15.3, α =0.05⇒tn−1,α/2 = t89,.025 =1.987

H0 : µ =21.7 versus HA : µ6=21.7

Rejection Region:|t|> t89,.025 =1.987


t =Y−µ0 S/sqrt(n) =18.8−21.7 15.3/p90 = −2.9 1.6128 =−1.798⇒|t|=1.798 < t89,.025 =1.987

Cannot reject H0 at 5% level. There is not sufficient evidence to support the average number of Type 2 fibers is different from 21.7.

what is the p-value?

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Answer #1

Hence we get-  P value = 0.0755

Hope this will help you. Thank you :)

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