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2 years ago

I WILL MARK BRAINLIST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

2 answers:

Nana76 [90]2 years ago

6 0

Answer:

a) 7 were 30 or younger

4 of ≤ 30 and 3 of ≥ 30 has no facial hair

b) as attached

c) c) 5/15= 1/3 ( 33.33%) of men older than 30 has facial hair, in total group

0r 5/8 = 0.625 (62.5%) of men older than 30 in his age group has facial hair

Step-by-step explanation:

a) b) shown in attached

c) 5/15= 1/3 ( 33.33%) of men older than 30 has facial hair, in total group

0r 5/8 = 0.625 (62.5%) of men older than 30 in his age group has facial hair

Kisachek [45]2 years ago

3 0

Answer:

(a) 30 & younger (no facial hair are 1,2,4,9. with facial hair are 10,12,14)

older than 30 (no facial hair are 3,4,8 . with facial hair are 6,7,11,13,15)

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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.

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$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

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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.