Please answer quick you will be rewarded handsomely!!!

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

You need a 45% alcohol solution. On hand, you have a 175 mL of a 25% alcohol mixture. You also have 70% alcohol mixture. How muc

Bezzdna [24]

175 mL at 25% concentration of alcohol contains 0.25 (175 mL) = 43.75 mL of alcohol. If v is the amount of the 70% solution that you use, then that amount contains 0.7v mL of alcohol.

Mixing these two yields a total volume of 175 mL + v, and it contains 43.75 mL + 0.7v alcohol. You want to end up with a concentration of 45%, which means the ratio of the amount of alcohol to the total volume needs to be 0.45:

(43.75 mL + 0.7v) / (175 mL + v) = 0.45

Solve for v :

43.75 mL + 0.7v = 0.45 (175 mL + v)

43.75 mL + 0.7v = 78.75 mL + 0.45v

0.25v = 35 mL

v = 140 mL

5 0

3 years ago

PLEASE HELP MEEEEEEEEEEEEEEEEEE!!!!!!!!

victus00 [196]

F = 41
E = 90
D = 49
from D to E is 55
then E to F also known as X = 36.08 or 41.51

6 0

3 years ago

Does anybody know the answer to this ?

melisa1 [442]

Answer:

Step-by-step explanation:

5 0

3 years ago

How to represent 0.5 on a decimal grid

Alina [70]

Answer:

put one and then put 0.5 in between 0 an 1

Step-by-step explanation:

7 0

2 years ago