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3 years ago

Translate the statement into a confidence interval for p. Approximate the level of confidence.

Mathematics

1 answer:

timurjin [86]3 years ago

6 0

Answer:

The interval is approximately a 98% confidence interval

Step-by-step explanation:

From the question : Error, E= 3% = 0.03, Total population, n=1000, number of people that are concerned that their taxes will be audited, p = 19% = 0.19

E^{2}=z_{\alpha/2}^{2}\cdot \frac{p(1-p)}{n}

0.03^{2}=z_{\alpha/2}^{2}\cdot \frac{0.19(1-0.19)}{1000}

z_{\alpha/2}^{2}=5.848

z_{\alpha/2}=2.418

Area right to 2.418 is 0.0078. So

\alpha/2=0.0078

Therefore \alpha=0.0156\approx 0.02

Thus, the interval is approximately a 98% confidence interval.

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Find the quotient 8/1 ÷ 3/5

den301095 [7]

13 1/3 is the answer, or in mixed numbers should be 40/3

5 0

3 years ago

Describe how you can tell that a fraction corresponds to a repeating decimal and not a terminating decimal

horrorfan [7]

When a fraction corresponds to a repeating decimal, it's denominator is 9 or 99.

For example, 4/11 = 36/99 = 0.36 with bar notation over 36.

Example #2: 2/3 = 6/9 = 0.666....
or also written as 0.6 with bar notation over 6.

You know a fraction responds to a terminating decimal because it can be simplified to a terminating decimal.

Example: 2/4 = 0.5

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3 years ago

A doctor is interested in people who have experienced an unexplained episode of vitamin D intoxication. She took a SRS of size 1

Mila [183]

Answer:

(a) (2.573, 3.167) is the 99% confidence interval for $\mu$ the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication.

(b) There is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

Step-by-step explanation:

If we have a random sample of size n from a normal distribution with mean $\mu$ and standard deviation $\sigma$ , then we know that $\bar{X}$ is normally distributed with mean $\mu$ and standard deviation $\sigma/n$ . Therefore we can use $(\bar{X}-\mu)/\frac{\sigma}{\sqrt{n}}$ as a pivotal quantity.

We have a sample size of n = 12. The sample mean is $\bar{x}$ = 2.87 mmol/l and the standard deviation is $\sigma = 0.40$ . The confidence interval is given by $\bar{x}\pm z_{\alpha/2}(\frac{\sigma}{\sqrt{n}})$ where $z_{\alpha/2}$ is the $\alpha/2$ th quantile of the standard normal distribution. As we want the 99% confidence interval, we have that $\alpha = 0.01$ and the confidence interval is $2.87\pm z_{0.005}(\frac{0.40}{\sqrt{12}})$ where $z_{0.005}$ is the 0.5th quantile of the standard normal distribution, i.e., $z_{0.005} = -2.5758$ . Then, we have $2.87\pm (-2.5758)(\frac{0.40}{\sqrt{12}})$ and the 99% confidence interval is given by (2.573, 3.167)

(b) We found a 99% confidence interval for the true mean calcium level of the population of people who experienced an unexplained episode of vitamin D intoxication. Therefore, there is a 99% probability that the average calcium level of all people who have experienced an unexplained episode of vitamin D intoxication is in the interval.

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4 years ago

Compare. Least to Greatest. SHOW UR WORK‼️

Natali5045456 [20]

2/4 is smaller
This is because 2/4 = 6/12
6/12 is smaller than 7/12
So therefore 2/4 is smaller

3 0

3 years ago

Read 2 more answers

Why does you not has?

frosja888 [35]

Answer:

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