A sample of 49 observations is selected from a normal population. The sample mean is 61, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.10 significance level.

t = (61-63)/(6/7) = -2*7/6 = -7/3 = -2.333

Tabular value for 1% (two tailed) is 1.6772 for 48 df

| t | =2.333 > 1.6772

we reject the null hypothesis.

And conclude that mean is not equal to 63

How about this question What is the p-value?

OK .

P( |t| >2.333) = 0.0239

p value =0.0239

since 0.1 >0.0239 or 0.0239 <0.1 we reject the null hypothesis.

p value =0.0239 it says wrong

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