HereSolutionNull hypotheis Y 05 Alternate hypothesis Y 5

Here

Null hypotheis : Y = 0.5

Alternate hypothesis : Y > .5

test statistic = ( y-bar - mean)/sd/sqrt(n)

= (0.5150 - 0.5)/0.12/sqrt(2500)

= 6.25

p value corresponding to t statistic and degrees of freedom n-1 = 0.00001

Since, p value < Significance level ( Assumed as 0.05)

We, reject the null hypothesis.

Conclusion : the mean of the data > 0.5

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99% confidence interval

= ( y bar - t(Alpha/2,n-1)*sd/sqrt(n) , y bar + t(Alpha/2,n-1)*sd/sqrt(n) )

Here, t(alpha/2, n-1) = t(0.005,9999) = 2.576

= ( 0.515 - 2.576*0.12/sqrt(10000) , 0.515 + 2.576*0.12/sqrt(10000))

= (0.5119 , 0.5181)