Public Health and Nutrition The U.S. Centers for Disease Control announced that

Question

Public Health and Nutrition The U.S. Centers for Disease Control announced that skin allergies have risen among U.S. children to 12.5%.22 Serious allergies can affect a child's education, sleep, and ordinary daily activities. Suppose 320 U.S. children are randomly selected and tested for skin allergies. Find the probability that the sample proportion of children with skin allergies is less than 0.10. Find the probability that the sample proportion of children with skin allergies is between 0.08 and 0.16. Suppose 150 U.S. children aged 5 to 9 are randomly selected and each is tested for skin allergies. Nine children are found to have skin allergies. Is there any evidence to suggest that children in this age group have a lower incidence rate of skin allergies? Justify your answer.

Explanation / Answer

Let n = number of childrens = 320

p = proportion = 12.5% = 0.125

np = 320*0.125 = 40

nq = 320*0.875 = 280

Both np and nq are > 10.

And we know that the distribution of sample proportion is normal with

mean = p = 0.125

sd = sqrt((p*q) / n) = sqrt((0.125*0.875) / 320) = 0.0185

a) Here we have to find P(p^ < 0.10).

Convert p^ = 0.10 into z-score.

z = (p^ - mean) / sd

z = (0.10 - 0.125) / 0.0185 = -1.35

That is now we have to find P(Z < -1.35).

This probability we can find by using EXCEL.

syntax is,

=NORMSDIST(z)

where z is test statistic value.

P(Z < -1.35) = 0.0881

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b) Here we have to find P(0.08 < p^ < 0.16).

Convert p^ = 0.08 and p^ = 0.16 into z-score.

z = (0.08 - 0.125) / 0.0185 = -2.43

z = (0.16 - 0.125) / 0.0185 = 1.89

That is now we have to find P(-2.43 < Z < 1.89).

P(-2.43 < Z < 1.89) = P(Z < 1.89) - P(Z < -2.43)

= 0.9708 - 0.0075

P(-2.43 < Z < 1.89) = 0.9634

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c) n = number of U.S. children aged 5 to 9 = 150

x = number of childrens are found to have skin allergies = 9

That is here we have to test the hypothesis that,

H0 : p = 0.125 Vs H1 : p < 0.125

where p is population proportion.

Assume alpha = level of significance = 5% = 0.05

This we can done using TI-83 calculator.

steps :

STAT --> 5: 1-PropZTest --> ENTER --> Input p0, x and n --> prop : select "<p0" --> ENTER --> Calculate --> ENTER

Output is :

z = -2.4071

P-value = 0.008

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that children in this age group have a lower incidence rate of skin allergies.

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