Compute the following without using a calculator. You must show your work to receive full credit.

Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average

aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

4 years ago

There are three boxes: one with two golden coins, one with two silver coins, and one with one golden coin and one silver coin. A

krek1111 [17]

Answer:

Step-by-step explanation:

From Bayes' theorem is stated mathematically as the following equation:[2]

{\displaystyle P(A\mid B)={\frac {P(B\mid A)\,P(A)}{P(B)}},}

where A and B are events and P(B) ≠ 0.

P(A) and P(B) are the probabilities of observing A and B without regard to each other.

P(A | B), a conditional probability, is the probability of observing event A given that B is true.

P(B | A) is the probability of observing event B given that A is true.

At this point, go through the attached file before you continue with part B.

Part B)

P(silver) = P(silver from SS)+P(silver from GS)

note P(SS)=P(GG)=P(GS) = 1/3

P(silver from SS) = 1

P(silver from GS) = 1/2

hence

P(Silver from SS) = 1/3

P(Silver from GS) = 1/3 *1/2

P(Silver) = 1/3*1+1/3*1/2

required probability = P(Silver from SS)/P(Silver) = 2/3

4 0

3 years ago

Gary earned a gross pay of $1,047.30 last week. Using the fact that Social Security tax is 6.2% of gross pay, determine the amou

Westkost [7]

To find the answer, we multiply 1,047.30 by 6.2% which is 0.062 in percentage.
1047.30 x 0.062 = 64.93
64.93 is being taken away.

3 0

3 years ago

Read 2 more answers

From a group of 12 students, we want to select a random sample of 4 students to serve on a university committee. How many combin

borishaifa [10]

Answer:

495 combinations of 4 students can be selected.

Step-by-step explanation:

The order of the students in the sample is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

How many combination of random samples of 4 students can be selected?

4 from a set of 12. So

$C_{n,x} = \frac{12!}{4!(8)!} = 495$

495 combinations of 4 students can be selected.

8 0

3 years ago

The prisms and two nets shown below:<br><br> Plzzz answer part A B and C for BRILLIANT ANSWER

Arlecino [84]

Part A: Net A is correct
Net B is incorrect because de triangular sides do not close the opening left in both sides.

Part B: AB=3 in., BC=5in., CD=8.6in.

Part C: The surface area of the prism is the area of the the big rectangle in the net + the area of the 2 triangles

Area of the big rectangle
8.6• ( 3+4+5)= 103.2 in ^2
Area of the triangles
If we get the 2 trangles together along their longest side we get another rectangle
3•4 =12 in^2

Surface area of prism is 103.2+12=115.2 in^2

5 0

3 years ago