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A frequenter of a pub had observed that the new barman poured in average 0.47 liters of beer into the glass with a standard d

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Answer #1

Solution :

\mu= 0.5

\bar x =0.47

\sigma =0.18

n = 45

This is the LEFT tailed test .

The null and alternative hypothesis is ,

H0 :  \mu = 0.5

Ha : \mu <0.5

Test statistic = z

= (\bar x - \mu ) / \sigma / \sqrt n

= (0.47-0.5) / 0.18 / \sqrt 45

= -1.12

P(z <-1.12) = 0.1314

P-value = 0.1314

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