Answer:
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) ![z_{crit}= 1.64](https://tex.z-dn.net/?f=%20z_%7Bcrit%7D%3D%201.64)
And the rejection zone would be ![z>1.64](https://tex.z-dn.net/?f=%20z%3E1.64)
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:
![z_{crit}= 1.64](https://tex.z-dn.net/?f=%20z_%7Bcrit%7D%3D%201.64)
And the rejection zone would be ![z>1.64](https://tex.z-dn.net/?f=%20z%3E1.64)
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