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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 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 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
and
, the unbiased point estimate for
is
, i.e., 699-682 = 17.
The standard error is given by , i.e.,
= 9.8198.
a. We want to test vs
(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
and the observed value is
. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.
b. The endpoints for a 95% confidence interval for is given by
, i.e.,
where
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
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.
We have that if we assume standard exponential growth, the equation of the population will be:
if we start counting from the moment that the population was 7000. We are given that 7000* 
=12000, namely that P(5)=12000 and we need to find 
. Since e^(10k)=e^(5k)*e^(5k), and since we can solve for e^(5k) from P(5), we have:
e^(5k)=12000/7000 and we can calculate P(10). P(10)=7000
= 20571 people.
e^(5k)=12000/7000 and we can calculate P(10). P(10)=7000