Respuesta :
Answer:
[tex]z=\frac{0.415 -0.37}{\sqrt{\frac{0.37(1-0.37)}{200}}}=1.318[/tex]
[tex]p_v =P(z>1.318)=0.0937[/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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is NOT improved (higher than 0.37)
Step-by-step explanation:
Data given and notation
n=200 represent the random sample taken
X=83 represent the students said say that they approve of the new parking arrangement
[tex]\hat p=\frac{83}{200}=0.415[/tex] estimated proportion of students said say that they approve of the new parking arrangement
[tex]p_o=0.37[/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 is higher than 0.37.:
Null hypothesis:[tex]p \leq 0.37[/tex]
Alternative hypothesis:[tex]p > 0.37[/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{0.415 -0.37}{\sqrt{\frac{0.37(1-0.37)}{200}}}=1.318[/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 right tailed test the p value would be:
[tex]p_v =P(z>1.318)=0.0937[/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 FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion is NOT improved (higher than 0.37)
Answer:
We conclude that the percentage of the school's students approved of the parking is less than or equal to 37%.
Step-by-step explanation:
We are given that a local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school's students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 of the over 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement.
Let p = % of the school's students approved of the parking after the change
Let Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 0.37 {means that the percentage of the school's students approved of the parking is less than or equal to 37% after the change}
Alternate Hypothesis, [tex]H_a[/tex] : p > 0.37 {means that the percentage of the school's students approved of the parking is more than 37% after the change}
The test statistics will be used here is One-sample proportion test;
T.S. = [tex]\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = % of the school's students approved of the parking in a survey of 200 students = [tex]\frac{83}{200}[/tex] = 0.415
n = sample of students = 200
So, test statistics = [tex]\frac{0.415 - 0.37}{\sqrt{\frac{0.415(1-0.415)}{200} } }[/tex]
= 1.292
Now, at 0.05 significance level the z table gives critical value of 1.6449. Since our test statistics is less than the critical value so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.
Therefore, we conclude that the percentage of the school's students approved of the parking is less than or equal to 37% after the change and principal's claim was not effective.
