Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The estimate of the average stopping distance of all model A tires with a 94% confidence interval is
[tex]132.94 < \mu < 144.86[/tex]
Step-by-step explanation:
From the question we see that
The sample size is n = 8
The population standard deviation is [tex]\sigma = 6.7[/tex]
Generally the sample mean is mathematically represented as
[tex]\= x = \frac{\sum x_i }{n}[/tex]
=> [tex]\= x = \frac{ 145 + 152 + \cdots + 133}{8}[/tex]
=> [tex]\= x = 138.9[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 94 ) \%[/tex]
=> [tex]\alpha = 0.04[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n - 1[/tex]
=> [tex]df = 8 - 1[/tex]
=> [tex]df = 7[/tex]
Because the n < 30 we will making use of the t table
Generally from the t distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 7[/tex] is
[tex]t_{\frac{\alpha }{2}, 7 } = 2.517 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2} , df} * \frac{\sigma }{\sqrt{n} }[/tex]
=> [tex]E = 2.517 * \frac{6.7 }{\sqrt{8} }[/tex]
=> [tex]E =5.96 [/tex]
Generally 94% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex]138.9 -5.96 < \mu < 138.9 + 5.96[/tex]
=> [tex]132.94 < \mu < 144.86[/tex]
