Answer: No, the percentage of the fleet out of compliance is not different from their initial thought.
Explanation:
Since we have given that
n = 22
x = 5
So,
![\hat{p}=(x)/(n)=(5)/(22)=0.23](https://img.qammunity.org/2020/formulas/mathematics/high-school/vi35egc61wkrz3xshhto6o36mjijlj69yg.png)
he company initially believed that 20% of the fleet was out of compliance. Is this strong evidence the percentage of the fleet out of compliance is different from their initial thought.
so, p = 0.2
Hypothesis would be
![H_0:p=\hat{p}\\\\H_a:p\\eq \hat{p}](https://img.qammunity.org/2020/formulas/mathematics/high-school/jqvg0xc3qqu3l4xbb7i584uy0g4x1up2oq.png)
So, the t test statistic value would be
![t=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}\\\\\\t=\frac{0.23-0.20}{\sqrt{(0.2* 0.8)/(22)}}\\\\\\t=(0.03)/(0.085)\\\\t=0.353](https://img.qammunity.org/2020/formulas/mathematics/high-school/ov5i2uhsougtemi85wqqafu8wvqv34j4n6.png)
Degree of freedom = df = n-1 = 22-1 =23
So, t{critical value} = 2.080
So, 2.080>0.353
so, we will accept the null hypothesis.
Hence, No, the percentage of the fleet out of compliance is not different from their initial thought.