Suppose that 11% of adolescent females in the United States have iron deficiency

Question

Suppose that 11% of adolescent females in the United States have iron deficiency. A researcher wants to test the theory that more than 11% of vegetarian adolescent females in the United States have an iron deficiency. To test this theory, a random sample of 100 vegetarian adolescent females is randomly selected and 15 are found to have an iron deficiency.

What is the p-value?

(a) 0.100

(b) 0.900    

(c) 0.131

(d) 0.869

We can't answer the question because the conditions aren't met since 100x0.11 < 15

Suppose the correct answer to (b) is 0.154 and = .05. Which of the following is the best conclusion?

1- The data provides statistical evidence that 11% of vegetarian adolescent females in the United States have an iron deficiency.

2- The data provides no statistical evidence that 11% of vegetarian adolescent females in the United States have an iron deficiency.    

3- The data provides no statistical evidence that more than 11% of vegetarian adolescent females in the United States have an iron deficiency.

4- The data provides no statistical evidence that 15% of vegetarian adolescent females in the United States have an iron deficiency.

5- The data provides no statistical evidence that 15% of these 100 vegetarian adolescent females in the United States have an iron deficiency

Explanation / Answer

Here, p = 15/100 = 0.15

Hypothesis:

H0 : P = .11

Ha : P > .11

Test statistic;

= sqrt[ P * ( 1 - P ) / n ]

= sqrt [(0.11 * 0.89) / 100]

= 0.031

z = (p - P) /

= (.15 - .11)/0.031 = 1.28

Now, we need to find p value by t statistic = 1.28

P value = 0.100273.

Answer is option A)

Here, p value = 0.154 and alpha = 0.05

It means p value is greater than 0.05 so we failed to reject the null hypothesis.

3- The data provides no statistical evidence that more than 11% of vegetarian adolescent females in the United States have an iron deficiency.

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