Question

An automobile insurance company claims that its rates for teenage drivers average $410 less per year than the same coverage from another company. 
In a random sample of 35 customers, the average savings was $390 per year, with a standard deviation of $55 per year. What is the z-value rounded to the nearest hundredth?  Is there enough evidence to reject the claim?​

Answer 1

Answer:

Z-value is -2.15.

Yes, It is enough evidence to Reject the claim.

Step-by-step explanation:

Given that,

Rate for teenage drivers average $410 less per year than the same coverage from another company. total number of customers are 35 and their average saving $390 per year. its standard deviation is $55.

To find :- What is the z-value rounded to the nearest hundredth ? is there enough evidence to reject the claim ?

so,  

               Z- value = $\frac{xbar- u}{\frac{\sigma}{\sqrt{n} } }$

Here,    $xbar = 390$,    $u= 410$,    $\sigma = 55$,   $n=55$

calculating the Z-value,

                Z-value = $\frac{390- 410}{\frac{55}{\sqrt{35} } }$

                             =   $\frac{-20\times5.9160}{55}$

                             = $-2.15$

Now, for rejecting the claim we need to find the P- value

(a) p value is used in hypothesis testing to find the support or reject of null hypothesis.

(b) the smaller p - value gives a strong evidence that you can reject the null hypothesis.

              P value = $P(z<-2.15) = 0.0157$

Here we get that the P value is smaller than $0.05$.    ($0.0157<0.05$)

Hence,

Yes, It is enough evidence to Reject the claim.