Question
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
Solutions
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 1 year agoRelated Solutions
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Previously, 5% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that...
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