HELLLL I’ll give Brainlist

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 the questions are in the picture above

svp [43]

Not sure if I'm right but I think it's 3(x - 6) (x^2 + 5x)

Step-by-step explanation:

3x^3 - 3x^2 - 90x

Apply GCF: 3 (x^3 - x^2 - 30)

Split 30 into -6 and 5

(x^3 - 6x^2) (5x^2 - 30x)

GCF of both: x^2 (x - 6) and 5x (x - 6)

DON'T FORGET TO CARRY THE 3

And your answer is 3 (x - 6) (x^2 + 5x)

5 0

3 years ago

A surveyor measures a road as being 69.3km long, however there is a 1℅ error in this measurement .What is the true length of the

Vaselesa [24]

Answer:

69.993

Step-by-step explanation:

Using the percentage error formula:

Percentage error = (true - measured value / true measurement ) * 100%

1% error in measurement

1% of 69.3

0.01 * 69.3 = 0.693

True measurement = error in measurement + measured value

True measurement = 0.693 + 69.3

Actual measurement = 69.993

Hence, actual measurement = 69.993.

5 0

2 years ago

Which number line shows the solution set to this inequality? -2x + 9 < x − 9

kotykmax [81]

Answer:

see below

Step-by-step explanation:

-2x + 9 < x − 9

Add 2x to each side

2x-2x + 9 < 2x+x − 9

9 < 3x-9

Add 9 to each side

9+9 < 3x

18 < 3x

Divide each side by 3

18/3 < 3x/3

6 < x

Open circle at 6 line going to the right

5 0

2 years ago

A cylindrical can has a volume of 2in^(3). Express the surface area of the can as a function of its radius,r.

noname [10]

$V=2 \text{ in}^3\\
A=2\pi r (r+h)\\
h=?\\\\
V=\pi r^2h=2\\
\pi r^2h=2\\
h=\dfrac{2}{\pi r^2}\\
A=2\pi r\left(r+\dfrac{2}{\pi r^2}\right)\\
\boxed{A=2\pi r^2+\dfrac{4}{r}}
 
$

7 0

2 years ago