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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

PLEASE HELP!!!!!!!!!!!!!! AND THANK YOU :-) 15 POINTS A concession-stand manager buys bottles of water and soda to sell at a foo

3241004551 [841]

The manager needs to buy 3375 water bottles and 1125 sodas and both will equal 4500. hope this helps

4 0

2 years ago

Read 2 more answers

The water tank in your school holds 20 liters of water. One day it was 34 full. That day, 14 liters of tank capacity was used up

Alex787 [66]

Answer:

20

Step-by-step explanation:

The water tank in the school hold 20 liters of water

One day it was 34 liters full

14 liters of water was used up

Therefore the quantity of water left can be calculated as follows

= 34-14

= 20

Hence 20 liters of water remains in the tank

6 0

2 years ago

Write an equation that could be used to find the value of x using the interior angles of ∆BDC, then solve it.

Alecsey [184]

9514 1404 393

Answer:

x = 16

Step-by-step explanation:

Either or both of the right triangles can be used to find x. Or, triangle ABC could be used. All numbers are assumed to be degrees.

Using ∆ABD

55 +90 +2x+3 = 180

2x = 32 . . . . . . subtract 148

x = 16

Using ∆BCD

50 +90 +2x+8 = 180

2x = 32 . . . . . . subtract 148

x = 16

Using ∆ABC

55 +(2x +3) +50 +(2x +8) = 180

4x = 64 . . . . . . . subtract 116

x = 16

6 0

2 years ago

I need help ASAP please giving brainliest!

klio [65]

900. It estimates bc one is like 288 and one is like 624 or sum.

7 0

2 years ago