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In: Statistics and Probability

​Previously, 44​% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that...

​Previously, 44​% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 44​% today. She randomly selects 120 pregnant mothers and finds that 22 of them smoked 21 or more cigarettes during pregnancy. Test the​ researcher's statement at the alpha equals α=0.1 level of significance

Find the​ P-value

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Expert Solution

Solution-

we have-

An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 44​%(0.44 = p) today. She randomly selects 120(N) pregnant mothers and finds that 22 (X) of them smoked 21 or more cigarettes during pregnancy.

Using given information hypothesis test is calculated

◆ Test Results and calculation-

The P-Value is 0.00001.

The result is significant at p < 0.10.

above test results that,

The null hypothesis Ho is rejected.

So , we have enough evidence to support researchers claim(statmstat) that the percentage of mothers who smoke 21 cigarette or more is less than 44%.

orchestra answered 2 years ago
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