Use the data below to test the following claim. A used car dealer says that the mean price of a three-year-old sport utility vehicle (in good condition) is $20,000. You suspect this claim is incorrect and find that a random sample of 22 similar vehicles has a mean price of $20,640 and a standard deviation of $1990. Is there enough evidence to reject the claim at a 0.05? Assume the population is normally distributed

a) Identify the claim and state He and Ha
b) Find the critical value(s) and identify the rejection region(s)
c) Find the standardized test statistic.
d) Decide whether to reject or fail to reject or fail to reject the null hypothesis
e) Interpret the decision in the context of the original claim.

(a) Null Hypothesis, : = $20,000   {means that the mean price of a three-year-old sport utility vehicle (in good condition) is $20,000}

Alternate Hypothesis, : $20,000   {means that the mean price of a three-year-old sport utility vehicle (in good condition) is different from $20,000}

(b) Critical values of t are -2.08 and 2.08 at 21 degree of freedom.

(c) The test statistics is 1.508.

(d) We fail to reject the null hypothesis.

(e) We conclude that the mean price of a three-year-old sport utility vehicle (in good condition) is $20,000.

Step-by-step explanation:

We are given that a random sample of 22 similar vehicles has a mean price of $20,640 and a standard deviation of $1990.

We have to test the claim of a used car dealer who says that the mean price of a three-year-old sport utility vehicle (in good condition) is $20,000.

Let = mean price of a three-year-old sport utility vehicle.

SO, Null Hypothesis, : = $20,000   {means that the mean price of a three-year-old sport utility vehicle (in good condition) is $20,000}

Alternate Hypothesis, : $20,000   {means that the mean price of a three-year-old sport utility vehicle (in good condition) is different from $20,000}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  =  ~

where,  = sample mean price of vehicles = $20,640

             s = sample standard deviation = $1,990

             n = sample of vehicles = 22

So, test statistics  =     ~

                               =  1.508

Now at 0.05 significance level, the t table gives critical values of -2.08 and 2.08 at 21 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject the null hypothesis.

Therefore, we conclude that the mean price of a three-year-old sport utility vehicle (in good condition) is $20,000.

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.