the us centers for disease control announced that skin allergies have risen amon

Question

the us centers for disease control announced that skin allergies have risen among US children to 12.5%. serious allergies can affect a child's education, sleep and ordinary daily activities. suppose 320 US children are randomly selected and tested for skin allergies. a)find the probability that the sample proportion of children with skin allergies is less than 0.10 b)find the probability that the sample proportion of children with skin allergies is between 0.08 and 0.16 c)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

Z-Score

The US centres for disease control announced that skin allergies have risen among US children to 12.5%. Serious allergies can affect a child's education, sleep and ordinary daily activities. Suppose 320 US children are randomly selected and tested for skin allergies.

We are given a population proportion = 12.5% = 0.125

Sample size = n = 320

Mean = np = 320*0.125 = 40

Here, q = 1 – p = 1 – 0.125 = 0.875

Standard deviation = sqrt(npq) = sqrt(320*0.125*0.875) = 5.9161

a)      Find the probability that the sample proportion of children with skin allergies is less than 0.10

We are given a population proportion = 12.5% = 0.125

Sample proportion = 0.10

Sample size = n = 320

Mean = np = 320*0.125 = 40

Here, q = 1 – p = 1 – 0.125 = 0.875

Standard deviation = sqrt (npq) = sqrt (320*0.125*0.875) = 5.9161

Here, we have to find P(sample proportion < 0.10)

X = 0.10*320 = 32

Now, we have to find P(X<32)

Z = (X – mean) / standard deviation

Z = (32 – 40) / 5.9161

Z = -1.35224

P(Z<-1.35224) = 0.088149

P(sample proportion < 0.10) = 0.088149

Required probability = 0.088149

b)      Find the probability that the sample proportion of children with skin allergies is between 0.08 and 0.16

We are given a population proportion = 12.5% = 0.125

Sample proportions = 0.08 and 0.16

Sample size = n = 320

Mean = np = 320*0.125 = 40

Here, q = 1 – p = 1 – 0.125 = 0.875

Standard deviation = sqrt (npq) = sqrt (320*0.125*0.875) = 5.9161

We have to find P(0.08<sample proportion<0.16)

X = 0.08*320 = 25.6

X = 0.16*320 = 51.2

P(25.6<X<51.2) = P(X<51.2) – P(X<25.6)

Find P(X<51.2)

Z = (51.2 – 40) / 5.9161 = 1.893139

P(Z<1.893139) = 0.97083

Find P(X<25.6)

Z = (25.6 – 40) / 5.9161 = -2.43404

P(Z<-2.43404) = 0.007466

P(25.6<X<51.2) = P(X<51.2) – P(X<25.6) = 0.97083 – 0.007466 = 0.963364

Required probability = 0.963364

c)       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

Here, we are given,

Sample size = n = 150

Sample proportion = 9/150 = 0.06

Here, we have to use the one sample z test for the population proportion.

The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: The children in the age group 5 to 9 have the incidence rate as 15% or 0.15.

Alternative hypothesis: Ha: The children in the age group 5 to 9 have the incidence rate lower than 15% or 0.15.

We assume the level of significance or alpha value as 0.05.

The test statistic formula is given as below:

Z = ( P – p ) / sqrt (pq/n)

P = sample proportion, p = population proportion, q = 1 – p and n = sample size

The test is given as below:

Z Test of Hypothesis for the Proportion

Data

Null Hypothesis            p =

0.15

Level of Significance

0.05

Number of Items of Interest

9

Sample Size

150

Intermediate Calculations

Sample Proportion

0.06

Standard Error

0.0292

Z Test Statistic

-3.0870

Lower-Tail Test

Lower Critical Value

-1.6449

p-Value

0.0010

Reject the null hypothesis

Here, we get the p-value as 0.0010 which is less than the given level of significance or alpha value 0.05, so we reject the null hypothesis that the children in the age group 5 to 9 have the incidence rate as 15% or 0.15.

This means, we concluded that the children in the age group 5 to 9 have the incidence rate lower than 15% or 0.15.

There is a sufficient evidence to suggest that children in the age group 5 to 9 have a lower incidence rate of skin allergies.

Z Test of Hypothesis for the Proportion

Data

Null Hypothesis            p =

0.15

Level of Significance

0.05

Number of Items of Interest

9

Sample Size

150

Intermediate Calculations

Sample Proportion

0.06

Standard Error

0.0292

Z Test Statistic

-3.0870

Lower-Tail Test

Lower Critical Value

-1.6449

p-Value

0.0010

Reject the null hypothesis

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