Explain which measure of center best describes a typical wage at this company:the mean or the median.

Khaleesi Herbert
2021-02-25
Answered

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sweererlirumeX

Answered 2021-02-26
Author has **91** answers

Here the mean is affected by the two higher salaries but the median does not affected by those values.

Therefore, median best describes a typical wage at this company.

Therefore, median best describes a typical wage at this company.

asked 2022-06-25

A company produces millions of 1-pound packages of bacon every week. Company specifications allow for no more than 3 percent of the 1-pound packages to be underweight. To investigate compliance with the specifications, the company’s quality control manager selected a random sample of 1,000 packages produced in one week and found 40 packages, or 4 percent, to be underweight.

Assuming all conditions for inference are met, do the data provide convincing statistical evidence at the significance level of $\alpha =0.05$ that more than 3 percent of all the packages produced in one week are underweight?

(A) Yes, because the sample estimate of 0.04 is greater than the company specification of 0.03.

(B) Yes, because the p-value of 0.032 is less than the significance level of 0.05.

(C) Yes, because the p-value of 0.064 is greater than the significance level of 0.05.

(D) No, because the p-value of 0.032 is less than the significance level of 0.05.

(E) No, because the p-value of 0.064 is greater than the significance level of 0.05.

The answer is (B) and I was trying to understand why. My calculation was:

${H}_{a}:p>0.03$

$z=\frac{\hat{p}-{p}_{0}}{\sqrt{\frac{{p}_{0}(1-{p}_{0})}{n}}}$

$z=\frac{0.04-0.03}{\sqrt{\frac{(0.03)(0.07)}{1000}}}$

But I get a ridiculously high number. I'm confused about how to get the p-value in this case.

A two-sided t-test for a population mean is conducted of the null hypothesis ${H}_{0}:\mu =100$. If a 90 percent t-interval constructed from the same sample data contains the value of 100, which of the following can be concluded about the test at a significance level of $\alpha =0.10$?

(A) The p-value is less than 0.10, and ${H}_{0}$ should be rejected.

(B) The p-value is less than 0.10, and ${H}_{0}$ should not be rejected.

(C) The p-value is greater than 0.10, and ${H}_{0}$ should be rejected.

(D) The p-value is greater than 0.10, and ${H}_{0}$ should not be rejected.

(E) There is not enough information given to make a conclusion about the p-value and ${H}_{0}$.

Here the answer is D, but again I am confused. How can I find the p-value in this case?

Assuming all conditions for inference are met, do the data provide convincing statistical evidence at the significance level of $\alpha =0.05$ that more than 3 percent of all the packages produced in one week are underweight?

(A) Yes, because the sample estimate of 0.04 is greater than the company specification of 0.03.

(B) Yes, because the p-value of 0.032 is less than the significance level of 0.05.

(C) Yes, because the p-value of 0.064 is greater than the significance level of 0.05.

(D) No, because the p-value of 0.032 is less than the significance level of 0.05.

(E) No, because the p-value of 0.064 is greater than the significance level of 0.05.

The answer is (B) and I was trying to understand why. My calculation was:

${H}_{a}:p>0.03$

$z=\frac{\hat{p}-{p}_{0}}{\sqrt{\frac{{p}_{0}(1-{p}_{0})}{n}}}$

$z=\frac{0.04-0.03}{\sqrt{\frac{(0.03)(0.07)}{1000}}}$

But I get a ridiculously high number. I'm confused about how to get the p-value in this case.

A two-sided t-test for a population mean is conducted of the null hypothesis ${H}_{0}:\mu =100$. If a 90 percent t-interval constructed from the same sample data contains the value of 100, which of the following can be concluded about the test at a significance level of $\alpha =0.10$?

(A) The p-value is less than 0.10, and ${H}_{0}$ should be rejected.

(B) The p-value is less than 0.10, and ${H}_{0}$ should not be rejected.

(C) The p-value is greater than 0.10, and ${H}_{0}$ should be rejected.

(D) The p-value is greater than 0.10, and ${H}_{0}$ should not be rejected.

(E) There is not enough information given to make a conclusion about the p-value and ${H}_{0}$.

Here the answer is D, but again I am confused. How can I find the p-value in this case?

asked 2021-11-18

Look at the normal curve below, and find $\mu ,\text{}\mu +\sigma$ , and $\sigma$

$$\begin{array}{|ccc|}\hline \mu & =& \\ \mu +\sigma & =& \\ \sigma & =& \\ \hline\end{array}$$

asked 2021-10-23

Let $X}_{1},{X}_{2},\dots ,{X}_{n$ be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables.

Find n such that$Pr\left(X>2000\right)\ge 0.95$ .

Find n such that

asked 2021-10-20

List 7 ways of representing data

asked 2022-06-13

Given that out of $1000$ individuals where $60$ use a drug. I am given the probability of a false positive is $0.009$ and the probability of a false negative is $0.10$. I'm trying to find the true positive and true negative.

Let $D$ be the even that the user uses a drug and ${D}^{C}$ the event the user is not a drug user. I know that $P(D)=0.06$ and $P({D}^{C})=0.994$ as well as $P(+|{D}^{c})=0.009$ and $P(-|D)=0.10$. So that means that $P(+|{D}^{C})=0.009\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}0.009\ast 990=8.91$ people. Also $P(-|D)=0.10\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}60\ast .10=6$ people. Does it follow that $P(-|{D}^{C})=981.09/990=0.991$ and $P(+|D)=54/60=0.90$?

Let $D$ be the even that the user uses a drug and ${D}^{C}$ the event the user is not a drug user. I know that $P(D)=0.06$ and $P({D}^{C})=0.994$ as well as $P(+|{D}^{c})=0.009$ and $P(-|D)=0.10$. So that means that $P(+|{D}^{C})=0.009\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}0.009\ast 990=8.91$ people. Also $P(-|D)=0.10\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}60\ast .10=6$ people. Does it follow that $P(-|{D}^{C})=981.09/990=0.991$ and $P(+|D)=54/60=0.90$?

asked 2022-05-30

According to the empirical rule, about 95% of the data lies within 2 standard deviations from the mean.

True or False?

asked 2022-03-08

Wilson interval with zero trials

I'm calculating Wilson score interval for some number of trials, so I have a question:

is it correct to assume this interval as [0 , 1] when you have zero trials?

I'm calculating Wilson score interval for some number of trials, so I have a question:

is it correct to assume this interval as [0 , 1] when you have zero trials?