What is the equation of the axis of symmetry of the graph of y + 3x – 6 = –3(x – 2)2 + 4?
2 answers:
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Answer:
1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.
2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].
Step-by-step explanation:
We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.
Let = <em>mean age of student cars.</em>
= <em>mean age of faculty cars.</em>
So, Null Hypothesis, :
{means that the student cars are younger than or equal to faculty cars}
Alternate Hypothesis, :
{means that the student cars are older than faculty cars}
(1) The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about the population standard deviations;
T.S. = ~
where, = sample mean age of student cars = 8 years
= sample mean age of faculty cars = 5.3 years
= sample standard deviation of student cars = 3.6 years
= sample standard deviation of student cars = 3.7 years
= sample of student cars = 110
= sample of faculty cars = 75
Also, =
= 3.641
So, <u><em>the test statistics</em></u> = ~
= 4.952
The value of t-test statistics is 4.952.
Since the value of our test statistics is more than the critical value of t, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.
Therefore, we support the claim that student cars are older than faculty cars.
(2) The 98% confidence interval for the difference between the two population means () is given by;
98% C.I. for () =
=
=
= [1.432, 3.968]
Here, the critical value of t at a 1% level of significance is 2.326.
Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].