A proportion, p, of people who are alcoholics is 60%. A random sample of 130 people...

A proportion, p, of people who are alcoholics is 60%. A random sample of 130 people is done, and 63 of them confused to be an alcoholic. Can we reject the hypothesis that .05 level of significance? Show work.

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Expert Solution

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p =0.60

Ha : p 0.60

= x / n = 63/130=0.485

P0 = 0.60

1 - P0 = 1-0.60=0.40

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.485-0.60/ [(0.60*0.40) /130 ]

= -2.68

P(z <-2.68 ) = 0.0038

P-value = 0.0038

= 0.05

P-value <

Reject the null hypothesis .

orchestra answered 1 year ago

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