Answer: No, the percentage of the fleet out of compliance is not different from their initial thought.
Step-by-step explanation:
Since we have given that
n = 22
x = 5
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{5}{22}=0.23[/tex]
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
[tex]H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}[/tex]
So, the t test statistic value would be
[tex]t=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\\\t=\dfrac{0.23-0.20}{\sqrt{\dfrac{0.2\times 0.8}{22}}}\\\\\\t=\dfrac{0.03}{0.085}\\\\t=0.353[/tex]
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.