Answer:
Yes there is sufficient evidence to reject the company's evidence
Explanation:
From the question we are told that
The sample size is n = 25
The mean is
The standard deviation is
![\sigma = 1.8 \ mpg](https://img.qammunity.org/2021/formulas/mathematics/high-school/32zfbogjcp6rxh55m3njxemvaxes8l6frj.png)
The z-score is z = -1.94
The null hypothesis is
![H_o : \mu = 6 \ mpg](https://img.qammunity.org/2021/formulas/mathematics/high-school/o4xnucxgzbiualf868y70u5ijzy78hpu8f.png)
The alternative hypothesis is
![H_a : \mu \\e 6](https://img.qammunity.org/2021/formulas/mathematics/high-school/8k5nz634tpawuzh3hv7qozbq9d9x8jo0uc.png)
Generally the p-value is mathematically evaluated as
![p-value = 2 * P( Z <-1.94)](https://img.qammunity.org/2021/formulas/mathematics/high-school/5rev8xdng15gv0v66jdlcy6ae7xhzmez7y.png)
From the z table the area under the normal curve to the left corresponding to -1.94 is
![P( Z <-1.94) = 0.02619](https://img.qammunity.org/2021/formulas/mathematics/high-school/picdav6bru90jtgagddc9op7pxksjjwdcu.png)
=>
![p-value = 2 * 0.02619](https://img.qammunity.org/2021/formulas/mathematics/high-school/io46kqiv5muonuy0a7kt20oek7x5d9fezr.png)
=>
![p-value = 0.052](https://img.qammunity.org/2021/formulas/mathematics/high-school/vsrkcfp68os6mz7jsl689rri7y2kul40br.png)
Let assume the level of significance is
![\alpha = 0.05](https://img.qammunity.org/2021/formulas/mathematics/college/445n2djo6b5zbv5df68kz5tjhh2puf9bol.png)
Hence the
this mean that
The decision rule is
Fail to reject the null hypothesis
The conclusion is
There is sufficient evidence to reject the company's evidence