Question

In a recent study, researchers found that 31 out of 150 boys aged 7-13 were overweight or obese. On the basis of this study can we conclude that more than 15% of the boys aged 7-13 in the sampled population are overweight or obese? Use a 5% level of significance. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value

Answer 1

1) Null hypothesis:
`p \leq 0.15`

Alternative hypothesis:
`p > 0.15`

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

3)
`z_(crit)= 1.64`

And the rejection zone would be
`z>1.64`

4) Calculate the statistic

`z=\frac{0.207 -0.15}{\sqrt{(0.15(1-0.15))/(150)}}=1.955`

5) Statistical decision

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

Explanation:

Data given and notation

n=150 represent the random sample taken

X=21 represent the boys overweight

`\hat p=(31)/(150)=0.207` estimated proportion of boy overweigth

`p_o=0.15` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

1) Concepts and formulas to use

We need to conduct a hypothesis in order to test the true proportion of boys obese is higher than 0.15.:

Null hypothesis:
`p \leq 0.15`

Alternative hypothesis:
`p > 0.15`

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)

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

3) Decision rule

For this case we need a value on the normal standard distribution who accumulates 0.05 of the area on the right tail and on this case this value is:

`z_(crit)= 1.64`

And the rejection zone would be
`z>1.64`

4) Calculate the statistic

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

`z=\frac{0.207 -0.15}{\sqrt{(0.15(1-0.15))/(150)}}=1.955`

5) Statistical decision

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15