PLZZ HELPP IM FAILLING THIS CLASS

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

3 years ago

The mean of the following data values is 32. <br> 19, 23, 35, 41, 42<br> A. True <br> B. False

wariber [46]

Answer:

False i think

Step-by-step explanation:

6 0

3 years ago

PLEASE SHOW WORK !!!

Fed [463]

For this, you have to understand the ratios of a 30,60,90 triangle. The ratio(in order) is x, x(sqrt 3), 2x (following the 30,60,90 pattern y'know). Well, we know the length of the hypotenuse, 4.2, so let's use ratios to figure the other sides out. The ratio of AC is x and the ratio of AB is 2x, so 2x/2=x=2.1. One side down. BC has a ratio of x(sqrt 3). We already know x so we can substitute it in. We get 2.1(sqrt 3). To calculate the area, use the formula 1/2(bh). Inputting the values in, we get 1/2(2.1*(2.1(sqrt 3)). This can be calculated to around 3.82. To calculate the perimeter, take the sum of the sides. By adding the sides together, we get about 9.98. Hope you can now understand how to do these kinds of questions.

3 0

2 years ago

98 POINTS!!!!! PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maslowich

Hello!

I believe the answer is C.

I hope it helps!

3 0

3 years ago

Read 2 more answers

Graph the solution of 2 ≥ 4 - v

vazorg [7]

Good evening ,

Answer:

2 ≤ v

Look at the photo below for the graph.

Step-by-step explanation:

2 ≥ 4 - v

2 - 4 ≥ - v

-2 ≥ -v

2 ≤ v.

:)

6 0

3 years ago