Question 1077022: Previously, 5% 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 5% today. She randomly selects
140 pregnant mothers and finds that
6 of them smoked 21 or more cigarettes during pregnancy. Test the researcher's statement at the α=0.05 level of significance.
What is the P value? I am having a hard time finding it
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Previously, 5% 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 5% today. She randomly selects
140 pregnant mothers and finds that
6 of them smoked 21 or more cigarettes during pregnancy. Test the researcher's statement at the α=0.05 level of significance.
What is the P value? I am having a hard time finding it
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Ho: p >= 0.05
Ha: p < 0.05 (claim)
Sample proportion:: 6/140 = 0.043
z(0.043) = (0.043-0.05)/sqrt[0.05*0.95/140] = -0.38
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P-value = P(z < -0.38) = normalcdf(-100,-0.38) = 0.35
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Conclusion:: Since the P-value is greater than 1%, fail
to reject Ho. The test results support the claim.
Cheers,
Stan H.
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