Khaleesi Herbert
2021-01-08
Answered

To find: The confidence interval for the mean population.

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SkladanH

Answered 2021-01-09
Author has **80** answers

The total size of the sample of the high school athletes is 233. The meanof the times they spend on the practicing is 1.6 h and the standard deviation is 0.5 h.
Obtain the z-value from the z-table of $99\mathrm{\%}$ confidence corresponding to the given mean and standard deviation is 2.576.
The confidence interval for the mean populationis calculated as,
$C=\text{}\stackrel{\u2015}{x}\text{}\pm \text{}z\text{}\times \text{}\frac{s}{\sqrt{n}}$

$=1.6\text{}\pm \text{}\left(2.576\right)\text{}\left(\frac{0.5}{\sqrt{233}}\right)$

$=1.6\text{}\pm \text{}0.0084$

$=1.516,\text{}1.684$
Thus, the confidence interval for the mean population is $1.516\text{}\le \text{}\mu \text{}\le \text{}1684.$

asked 2021-08-12

In a science fair project, Emily conducted an experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left hand, and then she asked the therapists to identify the selected hand by placing their hand just under Emily's hand without seeing it and without touching it. Among 358 trials, the touch therapists were correct 172 times. Complete parts (a) through (d).

a) Given that Emily used a coin toss to select either her right hand or her left hand, what proportion of correct responses would be expected if the touch therapists made random guesses? (Type an integer or a decimal. Do not round.)

b) Using Emily's sample results, what is the best point estimate of the therapists' success rate? (Round to three decimal places as needed.)

c) Using Emily's sample results, construct a

Round to three decimal places as needed - ?

asked 2021-03-09

In a study of the accuracy of fast food drive-through orders, Restaurant A had 298 accurate orders and 51 that were not accurate. a. Construct a 90% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part (a) to this 90% confidence interval for the percentage of orders that are not accurate at Restaurant B:

asked 2021-08-04

A random sample of 100 automobile owners in the state of Virginia shows that an automobile is driven on average 23,500 kilometers per year with a standard deviation of 3900 kilometers.

Assume the distribution of measurements to be approximately normal.

a) Construct a$99\mathrm{\%}$ confidence interval for the average number of kilometers an automobile is driven annually in Virginia.

b) What can we assert with$99\mathrm{\%}$ confidence about the possible size of our error if we estimate the average number of kilometers driven by car owners in Virginia to be 23,500 kilometers per year?

Assume the distribution of measurements to be approximately normal.

a) Construct a

b) What can we assert with

asked 2021-08-03

A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is $\sigma =15$

a) Compute the$95\mathrm{\%}$ confidence interval for the population mean. Round your answers to one decimal place.

b) Assume that the same sample mean was obtained from a sample of 120 items. Provide a$95\mathrm{\%}$ confidence interval for the population mean. Round your answers to two decimal places.

c) What is the effect of a larger sample size on the interval estimate?

Larger sample provides a-Select your answer-largersmallerItem 5 margin of error.

a) Compute the

b) Assume that the same sample mean was obtained from a sample of 120 items. Provide a

c) What is the effect of a larger sample size on the interval estimate?

Larger sample provides a-Select your answer-largersmallerItem 5 margin of error.

asked 2021-05-18

Two random variables X and Y with joint density function given by:

Find

asked 2022-04-22

In 2010, a medical research group reported the results of an experiment to evaluate the effectiveness of acupuncture to treat a chronic intestinal condition. A group of volunteers with the chronic intestinal condition agreed to participate in the experiment and be randomly assigned to either a true acupuncture treatment or a placebo treatment. The placebo treatment mimicked the application of acupuncture, but no needle penetrated the skin. Random assignment resulted in 78 subjects receiving acupuncture and 75 subjects receiving the placebo treatment. After receiving 6 treatments over the course of 3 weeks, patients were asked to report whether they had experienced a reduction in the chronic intestinal condition. The table summarizes the data from the study, with expected cell counts in parentheses.

$\begin{array}{|cccc|}\hline & \text{Yes}& \text{No}& \text{Total}\\ \text{Acupuncture}& 41(37.2)& 37(40.8)& 78\\ \text{Placebo treatment}& 32(35.8)& 43(39.2)& 75\\ \text{Total}& 73& 80& 153\\ \hline\end{array}$

Which of the following is true about the chi-square test for homogeneity?

A)The number of subjects randomly assigned to each treatment is not the same; therefore, it is not appropriate to use a chi-square test for homogeneity across treatment groups.

B)Volunteers do not constitute a random sample from the population of all patients with the chronic intestinal condition; therefore, it is not appropriate to use a chi-square test for homogeneity across treatment groups.

C)Volunteers with the chronic intestinal condition were randomly assigned to each treatment, so the independence condition has been met.

D)Not all of the observed cell counts are large enough to satisfy the conditions for applying the chi-square test of homogeneity.

E)Not all of the expected cell counts are large enough to satisfy the conditions for applying the chi-square test for homogeneity.

Which of the following is true about the chi-square test for homogeneity?

A)The number of subjects randomly assigned to each treatment is not the same; therefore, it is not appropriate to use a chi-square test for homogeneity across treatment groups.

B)Volunteers do not constitute a random sample from the population of all patients with the chronic intestinal condition; therefore, it is not appropriate to use a chi-square test for homogeneity across treatment groups.

C)Volunteers with the chronic intestinal condition were randomly assigned to each treatment, so the independence condition has been met.

D)Not all of the observed cell counts are large enough to satisfy the conditions for applying the chi-square test of homogeneity.

E)Not all of the expected cell counts are large enough to satisfy the conditions for applying the chi-square test for homogeneity.

asked 2022-07-08

Hence, $v(6,000)<v(4,000)+v(2,000)$ and $v(-6,000)>v(-4,000)+v(-2,000)$. These preferences are in accord with the hypothesis that the value function is concave for gains and convex for losses.

What this means and by how we can know if these align with convex and concave functions?

What this means and by how we can know if these align with convex and concave functions?