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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?​

1 Answer

4 votes

Answer:

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.

User Frantz
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