Question

Abnormalities In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation’s chil-dren. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the inci-dence of abnormalities. A recent study examined 384 children and found that 46 of them showed signs of an abnormality. Is this strong evidence that the risk has increased?

a) Write appropriate hypotheses.
b) Check the necessary assumptions and conditions.

Answer 1

Null hypothesis:
`p\leq 0.05`

Alternative hypothesis:
`p>0.05`

The conditions and requirements are explained on detail below.

Explanation:

1) Data given and notation n

n=384 represent the random sample taken

X=46 represent the children with abnormalities in the sample

`\hat p=(46)/(384)=0.120` estimated proportion of children with abnormalities in the sample

`p_o=0.05` is the value that we want to test

`\alpha` represent the significance level (no given)

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

p= proportion of children with congenital abnormalities

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion of children with congenital abnormalities exceeds 5%. :

Null hypothesis:
`p\leq 0.05`

Alternative hypothesis:
`p>0.05`

We assume that the proportion follows a normal distribution.

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly (different,higher or less) from a hypothesized value
`p_o`.

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

`np_o =384*0.05=19.2>10`

`n(1-p_o)=384*(1-0.05)=364.8>10`

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.120 -0.05}{\sqrt{(0.05(1-0.05))/(384)}}=6.29`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level is not provided usually is
`\alpha=0.05`. The next step would be calculate the p value for this test.

Since is a one side test the p value would be:

`p_v =P(z>6.29)=1.59x10^(-10)`

Based on the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of children with abnormalities exceeds 0.05 or 5% .