1. In a study of crime, the FBI found that 13.2% of all Americans had been victims of crime during a one-year period. This result was based on a sample of 1,105 adults. (a) Estimate the percentage of US adults who were victims at the 90% confidence level. (b) Estimate the percentage of US adults who were victims at the 99% confidence level.

Answers

Answer 1

Answer:

(a) The estimate the percentage of US adults who were victims at the 90% confidence level is (12%, 15%).

(b) The estimate the percentage of US adults who were victims at the 99% confidence level is (11%, 16%).

Explanation:

The percentage of Americans who had been victims of crime during a one-year period is, [tex]\hat p=0.132[/tex].

The sample selected was of size, n = 1105.

The (1 - α)% confidence interval for the population proportion is:

[tex]CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

(a)

The confidence level is 90%.

The critical value of z for 90% confidence level is, z = 1.645.

Compute the 90% confidence interval for the percentage of US adults who were victims as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

    [tex]=0.132\pm 1.645\times\sqrt{\frac{0.132(1-0.132)}{1105}}\\\\=0.132\pm 0.0168\\\\=(0.1152, 0.1488)\\\\\approx (0.12, 0.15)[/tex]

Thus, the estimate the percentage of US adults who were victims at the 90% confidence level is (12%, 15%).

(b)

The confidence level is 99%.

The critical value of z for 99% confidence level is, z = 2.58.

Compute the 99% confidence interval for the percentage of US adults who were victims as follows:

[tex]CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

    [tex]=0.132\pm 2.58\times\sqrt{\frac{0.132(1-0.132)}{1105}}\\\\=0.132\pm 0.0263\\\\=(0.1057, 0.1583)\\\\\approx (0.11, 0.16)[/tex]

Thus, the estimate the percentage of US adults who were victims at the 99% confidence level is (11%, 16%).


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40 points! SAT Math help

Answers

Answer: y = 4

Explanation:

Since line a and line b are parallel, they will have the same slope.

slope (m) for line a is:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{5-1}{4-(-2)}=\dfrac{4}{6}\quad \rightarrow \quad m=\large\boxed{\dfrac{2}{3}}[/tex]

Next, use the Point-Slope formula on line b to find y:

[tex]y_2-y_1=m(x_2-x_1)\\\\(x_1, y_1)=(-4,-4)\qquad (x_2,y_2)=(8.y)\qquad m=\dfrac{2}{3}\\\\\\y-(-4)=\dfrac{2}{3}\bigg(8-(-4)\bigg)\\\\\\y+4=\dfrac{2}{3}(12)\\\\\\y+4=2(4)\\\\\\y+4=8\\\\\\\large\boxed{y=4}[/tex]

salaries of 45 college graduates who took a statistic course in college have a mean of 68,500 assuming a standard deviation, of 10,990, construct 95% confidence interval for estimating the population mean mu​

Answers

Explanation:

We are given:

sample size= n = 45

sample mean = x = 61,300

sample standard deviation =18,246

Since the population standard deviation is not known we will use t distribution to find the confidence interval.

Confidence interval = 99%

Degrees of freedom= n - 1 = 45 – 1 = 44

Critical t value = 2.692

The 99% confidence interval will be:[tex](x-t_{critical}* \frac{s}{ \sqrt{n} }, x+t_{critical}* \frac{s}{ \sqrt{n} }) \\ \\ (61300-2.692* \frac{18246}{ \sqrt{45} }, 61300+2.692* \frac{18246}{ \sqrt{45} }) \\ \\ (53978,68622)[/tex]

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53978 to 68622

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Answers

Answer:

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Explanation:

Given that ∠A and ∠B are supplementary angles so when both angles are added up, it will form 180°. ∠A is twice the measure of ∠B :

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Answers

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