Asked by Keron Ashley on Jul 09, 2024
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A recent study used umbilical cord blood to test for 25 hydroxyvitamin D,which is an indicator of vitamin D status of the baby.It was reported that 65% of babies tested were deficient in vitamin D in spite of the fact that the mothers consumed vitamin D supplements during pregnancy.A researcher in a northern region felt that this percentage was too high for this region because with the reduced hours of sunshine during winter months,pregnant women tended to use higher doses of supplements to compensate.A sample of 125 newborns was tested,and 72 were declared to be deficient in vitamin D.What is the P-value and the conclusion using = 0.05?
A) P = 0.042;do not reject H0 since the P-value is less than the given level.
B) P = 0.047;do not reject H0 because it is significant at the given level.
C) P = 0.084;the result is not significant at the = 0.05 level.
D) P = 0.041;reject H0 since the P-value is less than the given level.
E) We are unable to calculate the P-value or reach a conclusion with the information given.
Vitamin D Supplements
Nutritional products designed to increase intake of vitamin D, a nutrient essential for bone health and immune function.
P-value
A statistical measure that helps to determine the significance of results obtained in a hypothesis test.
- Comprehend the importance of the significance level and P-value within hypothesis testing scenarios.
Verified Answer
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Aleyssa De JesusJul 15, 2024
Final Answer :
D
Explanation :
To determine the P-value, we need to conduct a hypothesis test. Our null hypothesis (H0) is that the proportion of newborns deficient in vitamin D is equal to or less than 65% (the reported percentage). Our alternative hypothesis (Ha) is that the proportion is greater than 65%.
We can use a one-sample proportion test and calculate the test statistic z as follows:
z = (p̂ - P) / √[P(1-P)/n]
where p̂ is the sample proportion of newborns deficient in vitamin D, P is the hypothesized proportion (0.65), and n is the sample size.
Plugging in the values from the problem, we get:
z = (0.576 - 0.65) / √[(0.65)(0.35)/125] = -1.784
Using a standard normal distribution table, we find that the P-value is approximately 0.041.
Since the P-value (0.041) is less than the given level of significance (0.05), we reject H0 and conclude that there is evidence to suggest that the proportion of newborns deficient in vitamin D in this northern region is higher than 65%. Therefore, the best choice is D.
We can use a one-sample proportion test and calculate the test statistic z as follows:
z = (p̂ - P) / √[P(1-P)/n]
where p̂ is the sample proportion of newborns deficient in vitamin D, P is the hypothesized proportion (0.65), and n is the sample size.
Plugging in the values from the problem, we get:
z = (0.576 - 0.65) / √[(0.65)(0.35)/125] = -1.784
Using a standard normal distribution table, we find that the P-value is approximately 0.041.
Since the P-value (0.041) is less than the given level of significance (0.05), we reject H0 and conclude that there is evidence to suggest that the proportion of newborns deficient in vitamin D in this northern region is higher than 65%. Therefore, the best choice is D.
Learning Objectives
- Comprehend the importance of the significance level and P-value within hypothesis testing scenarios.