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

Answer:

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  

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

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

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=21 represent the boys overweight

$\hat p=\frac{31}{150}=0.207$ 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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