Respuesta :
Answer:
a) Null hypothesis:[tex]p=0.5[/tex]
Alternative hypothesis:[tex]p \neq 0.5[/tex]
[tex]z=\frac{0.242-0.5}{\sqrt{\frac{0.5(1-0.5)}{484}}}=-11.352[/tex]
[tex]p_v =2*P(z<-11.352) \approx 0[/tex]
So the p value obtained was a very low value 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of new engineering graduates were planning graduate study differs from 0.5.
b) If we find the confidence interval for the real population proportion and we don't got the 0.5 in the interval we can reject the null hypothesis that p =0.5
[tex]0.242 - 1.96 \sqrt{\frac{0.242(1-0.242)}{484}}=0.204[/tex]
[tex]0.242 + 1.96 \sqrt{\frac{0.242(1-0.242)}{484}}=0.280[/tex]
And the 95% confidence interval would be given (0.204;0.280). So we see that the upper bound for the confidence interval is <0.5 so we can reject the nu;l hypothesis that p=0.5 at 5% of significance.
Step-by-step explanation:
Data given and notation
n=484 represent the random sample taken
X=117 represent the new engineering graduates were planning graduate study.
[tex]\hat p=\frac{117}{484}=0.242[/tex] estimated proportion of new engineering graduates were planning graduate study.
[tex]p_o=0.5[/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 the true proportion of new engineering graduates were planning graduate study is 0.5 or no:
Null hypothesis:[tex]p=0.5[/tex]
Alternative hypothesis:[tex]p \neq 0.5[/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{0.242-0.5}{\sqrt{\frac{0.5(1-0.5)}{484}}}=-11.352[/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 bilateral test the p value would be:
[tex]p_v =2*P(z<-11.352) \approx 0[/tex]
So the p value obtained was a very low value 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of new engineering graduates were planning graduate study differs from 0.5.
Part b
If we find the confidence interval for the real population proportion and we don't got the 0.5 in the interval we can reject the null hypothesis that p =0.5
The confidence interval would be given by this formula
[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=1.96[/tex]
And replacing into the confidence interval formula we got:
[tex]0.242 - 1.96 \sqrt{\frac{0.242(1-0.242)}{484}}=0.204[/tex]
[tex]0.242 + 1.96 \sqrt{\frac{0.242(1-0.242)}{484}}=0.280[/tex]
And the 95% confidence interval would be given (0.204;0.280). So we see that the upper bound for the confidence interval is <0.5 so we can reject the nu;l hypothesis that p=0.5 at 5% of significance.
