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3 years ago

7

jim’s percentile rank in a class of 50 students was the 92nd. what was his rank in the class (1st = best, 2nd = second best, etc

.)?

1 answer:

dezoksy [38]3 years ago

4 0

Jim is ranked third best in his class of 50 students.

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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

4 years ago

Find the value of p for which x = -2, y = -1 is a solution of linear equation 5x + 2py = 2p.

Rom4ik [11]

I think p=4
How I solved it:
5(-2)+2p+2(-1)=2p
-10+2p+-2=2p So cross out last
-2p -2p 2p:)
= -10+2p+-2
+10 +10 Cross out the 10
= 2p=8
= divide both sides by 2p
= 2p/2= 8/2
= cross out the 2.
= which leaves
= p=4 we get 4 because we divide 8 to 2:)

I don’t know if I’m right but yea...

8 0

3 years ago

A total of 243 adults and children are at a movie theater. There are 109 more adults than children in the theater. If eh represe

Marina86 [1]

The system should look like this:

eh + b = 243
eh - b = 109

6 0

4 years ago

Add (13x – 4) and (–6x + 15)

kompoz [17]

(13x – 4) + (–6x + 15) =

13x - 4 - 6x + 15 =

7x + 11

4 0

3 years ago

Why does 498 * 8.32 equal 4 1/4 3.36

Svetradugi [14.3K]

Answer:

The answer is 4143. 36 but I don't know why there are fractions

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers