Respuesta :
Answer:
a) [tex]z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31[/tex]
[tex]p_v =P(Z<-1.31)=0.095[/tex]
If we compare the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.
b) We can use a 95 percent upper confidence interval.
On this case we want a interval on this form : [tex](-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})[/tex]
So the critical value would be on this case [tex]Z_{\alpha}=1.64[/tex] and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"
We found the upper limit like this:
[tex]0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026[/tex]
And the interval would be: [tex](-\infty,0.03026)[/tex]
And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03
Step-by-step explanation:
Part a
Data given and notation
n=500 represent the random sample taken
X=10 represent the number of objects rejected
[tex]\hat p=\frac{10}{500}=0.02[/tex] estimated proportion of objects rejected
[tex]p_o=0.03[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns.:
Null hypothesis:[tex]p=0.03[/tex]
Alternative hypothesis:[tex]p < 0.03[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.02 -0.03}{\sqrt{\frac{0.03(1-0.03)}{500}}}=-1.31[/tex]
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 provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a one tailed left test the p value would be:
[tex]p_v =P(Z<-1.31)=0.095[/tex]
If we compare the p value obtained and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of rejected items is less than 0.03.
Part b
We can use a 95 percent upper confidence interval.
On this case we want a interval on this form : [tex](-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})[/tex]
So the critical value would be on this case [tex]Z_{\alpha}=1.64[/tex] and we can use the following excel code to find it: "=NORM.INV(1-0.05,0,1)"
We found the upper limit like this:
[tex]0.02+1.64\sqrt{\frac{0.02 (1-0.02)}{500}}=0.03026[/tex]
And the interval would be: [tex](-\infty,0.03026)[/tex]
And since our value (0.02) is contained in the interval We fail to reject the hypothesis that p=0.03
