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

Z-value is -2.15.

Yes, It is enough evidence to Reject the claim.

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 =
`(xbar- u)/((\sigma)/(√(n) ) )`

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

calculating the Z-value,

Z-value =
`(390- 410)/((55)/(√(35) ) )`

=
`(-20*5.9160)/(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.