Question

The following 4 questions are related. After seeing countless commercials claiming one can get cheaper car insurance from an online company, a local insurance agent was concerned that he might lose some customers. To investigate, he randomly selected profiles for 10 of his clients and checked online price quotes for their policies. Then, based on paired data on the 10 clients (the price he offered them and the corresponding online quote) and using Excel’s t-test for paired data, he obtained t-statistic = 0.709 p-value = 0.248 Note that the above is based on the difference between his price and online quotes.

Question 30 Not yet answered Points out of 1.00 Flag question Question text What are the null and alternative hypotheses, respectively? Select one: a. μd<=0 and μd > 0 b. μd>0 and μd <= 0

Question 31 Not yet answered Points out of 1.00 Flag question Question text At α=0.1, Select one: a. He can reject the null in favor of the alternative hypothesis b. He cannot reject the null hypothesis

Question 32 Not yet answered Points out of 1.00 Flag question Question text What degrees of freedom were used in the test? Select one: a. 8 b. 9 c. 10 d. 7

Question 33 Not yet answered Points out of 1.00 Flag question Question text Based on his evidence, should he be worried about his online competitors? Select one: a. Yes b. No

Answer 1

The null hypothesis is that there is no difference in prices between the agent and online quotes, and the alternative hypothesis suggests the agent's prices are higher. The agent cannot reject the null hypothesis at α=0.1, and with 9 degrees of freedom, there's no evidence that he should be worried about online competition.

Step-by-step explanation:

The null hypothesis (H0) and alternative hypothesis (Ha) respectively address whether the local agent's prices are higher than online prices. In this case, since the t-test for paired data is typically looking for differences, the null hypothesis could be that there is no difference in price (μd = 0). However, the choices given suggest it's interested in the direction of the difference, which implies:

• a. μd≤0 and μd > 0

For Question 31, given the p-value of 0.248, which is higher than the alpha level α=0.1, we cannot reject the null hypothesis. Thus, the agent cannot conclude that his prices are higher than online prices at the 10% significance level.

Question 32 involves degrees of freedom, which for a paired t-test is n-1 (where n is the number of pairs), resulting in 9 degrees of freedom for 10 pairs.

With the provided evidence, the insurance agent should not be worried about online competitors because there is no statistical evidence at α=0.1 to show that online quotes are cheaper based on the paired t-test results.

Answer 2

Question 30

Null hypothesis:
`\mu_y- \mu_x =\mu_d \leq 0`

Alternative hypothesis:
`\mu_y -\mu_x =\mu_d >0`

So the best option is:

a. μd<=0 and μd > 0

Question 31

So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is lower or equal than 0.

Then the best option is:

b. He cannot reject the null hypothesis

Question 32

`df=n-1=10-1=9`

b. 9

Question 33

Since we FAIL to reject the null hypothesis we don't have enough evidence to conclude that we don't have significant differences at 1% of significance.

Step-by-step explanation:

Previos concepts

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Question 30

The system of hypothesis for this case are:

Null hypothesis:
`\mu_y- \mu_x =\mu_d \leq 0`

Alternative hypothesis:
`\mu_y -\mu_x =\mu_d >0`

So the best option is:

a. μd<=0 and μd > 0

Question 31

We have the statistic already calculate t=0.709

The next step is calculate the degrees of freedom given by:

`df=n-1=10-1=9`

Now we can calculate the p value, since we have a right tailed test the p value is given by:

`p_v =P(t_((9))>0.709) =0.248`

And we can calculate it using the following excel code:

"=1-T.DIST(0.709,9,TRUE)"

So the p value is higher than any significance level given, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is lower or equal than 0.

Then the best option is:

b. He cannot reject the null hypothesis

Question 32

`df=n-1=10-1=9`

b. 9

Question 33

b. No

Since we FAIL to reject the null hypothesis we don't have enough evidence to conclude that we don't have significant differences at 1% of significance.