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9. Use a significance level of a = 0.05 to test the claim that p = 32.6. The sample consists of 15 scores for which x = 40.8
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Answer #1

a) Hypothesis : H_0:\mu=32.6 ( Claim ) VS H_a:\mu\neq32.6

b) Signiifcance level== 0,05

c) The test is two-tailed test.

because yje alternative hypothesis is not equal to type

d) The critical value is ,

t_{n-1,\alpha/2}=t_{15-1,0.05/2}=\pm 2.145 ; From t-table

e) The test statistic is ,

t_{stat}=\frac{\bar{X}-\mu_0}{s/\sqrt{n}}=\frac{40.8-32.6}{7.9/\sqrt{15}}=4.021

the p-value is ,

p-value=P(t_{d.f.}>|t_{stat}|)=P(t_{14}>4.0201)=0.0013

The Excel function is , =TDIST(4.0201,14,2)

f) Decision : Here , p-value=0.0013 < = 0,05

Therefore , reject Ho.

Conclusion : Hence , there is not sufficient evidence to support the claim.

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