Jack has 3 quarters, 2 dimes, and one penny. how much money does jack have in all? show your work.

saveliy_v [14]

The best way to find your answer is to divide 193 by 400 to get .4825 and to make sure if it is the answer you take 400 and multiply .4825 and it will be 193 so .4825 = 48.25%

3 0

3 years ago

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.