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

1) Null hypothesis:  

Alternative hypothesis:  

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

3)

And the rejection zone would be

4) Calculate the statistic  

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

estimated proportion of boy overweigth

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:  

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

(1)  

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

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:

And the rejection zone would be

4) Calculate the statistic  

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

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

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.