Question

In: Math

A sample of 200 science professor annual salaries yielded a mean value of $60,000. If the...

A sample of 200 science professor annual salaries yielded a mean value of $60,000. If the variance of population salary is $225 million, answer the following questions at 99% confidence The Point Estimate is ..... The Standard error is ..... The Margin of Error is ..... The Critical Statistic is ...... The population mean is estimated to range from .......... to ........ in ...........% of the time

Expert Solution

Given that, sample size (n) = 200

sample mean = $60000 and

population variance = $225 million

1 million = 1000000

=> population standard deviation

standard error (SE) =

A 99% confidence level has significance level of 0.01 and critical value is,

Margin of error (E) is,

The 99% confidence interval for population mean is,

Hence,

The point estimate is $60000.

The standard error is 1060.6602.

The margin of error is 2731.20.

The critical statistic is 2.575

The population mean is estimated to range from 57268.80 to 62731.20 in 99% of the time.

Note : if you want answers rounded to nearest integer then,

The standard error is 1061

The margin of error is 2732

The critical statistic is 2.575

The population mean is estimated to range from 57268 to 62732 in 99% of the time.

milcah answered 2 months ago

. Thenutritionlabelonabagofpotatochipssaysthataoneounceservingofpotatochipshas130calories. A random sample of 35 bags yielded a sample mean of 134 calories with...

. Thenutritionlabelonabagofpotatochipssaysthataoneounceservingofpotatochipshas130calories. A random sample of 35 bags yielded a sample mean of 134 calories with a standard deviation of 17 calories. Is there signiﬁcant evidence that the nutrition label does not provide an accurate measure of calories in the bags of potato chips? Answer this question using the hypothesis testing framework. Explicitly give the p-value for your test.

A data set yielded the following information: The sample mean is 45. The sample standard deviation...

A data set yielded the following information: The sample mean is 45. The sample standard deviation is 17.1. (a) According to the z-score criterion, how small does a value in this data set have to be in order to be called an outlier from below? (b) According to the z-score criterion, how large does a value have to be in order to be called an outlier from above?

An economic surveys 31 adults living is Townsburg about their salaries. The mean annual salaries for...

An economic surveys 31 adults living is Townsburg about their salaries. The mean annual salaries for those surveyed was found to be x̄ = $47,782, with a standard deviation of $2,488. a) Find a 90% confidence interval for the true mean annual salary of adults living in Townsburg. b) Provide the margin of error of the interval as your answer. Round your answer to the nearest dollar.

8) An economic surveys 35 adults living is Townsburg about their salaries. The mean annual salaries...

8) An economic surveys 35 adults living is Townsburg about their salaries. The mean annual salaries for those surveyed was found to be x̄ = $48,240, with a standard deviation of $2,794. a) Find a 90% confidence interval for the true mean annual salary of adults living in Townsburg. b) Provide the margin of error of the interval as your answer. Round your answer to the nearest dollar.

The following is a random sample of the annual salaries of high school counselors in the...

The following is a random sample of the annual salaries of high school counselors in the United States. Assuming that the distribution of salaries is approximately normal, construct a 90% confidence interval for the mean salary of high school counselors across the United States. Round to the nearest dollar. $55,250, $46,540, $42,120, $58,740, $38,010, $43,650, $65,640

A random sample of 20 recent weddings in a country yielded a mean wedding cost of...

A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,327.16 Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8,100 Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, muμ, of all recent weddings in this country. The 95% confidence interval is from ????? (Round to the nearest cent as needed.)

The accompanying table shows the annual salaries, in thousands of dollars, earned by a sample of...

The accompanying table shows the annual salaries, in thousands of dollars, earned by a sample of individuals who graduated with MBAs in the years 2012 and 2013. Perform a hypothesis test using α equals= 0.05 to determine if the median salary for 2012 MBA graduates is lower than 2013 MBA graduates. Salaries (in thousands of dollars) for 2012 MBA graduates 92.6 66.1 60.6 85.5 92.8 44.7 84.4 81.1 44.2 94.3 66.6 61.6 94.6 81.6 Salaries (in thousands of dollars) for...

The following set of data represents the distribution of annual salaries of a random sample of...

The following set of data represents the distribution of annual salaries of a random sample of 100managers in a large multinational company: Salary range (£` 000' ) Managers 20 but under 25 25 but under 30 30 but under 35 35 but under 40 40 but under 45 45 but under 50 5 10 25 35 25 5 Calculate the mean and standard deviation. [5 Marks] The company chairman claims that the managers in the company earn on average annual...

A random sample of 49 colleges yielded a mean cost of college education of $30,500 per...

A random sample of 49 colleges yielded a mean cost of college education of $30,500 per year. Assume that the population standard deviation is $3000 and college costs are normally distributed. a.) Calculate the standard error of the mean. Round to the hundredth. b.) Calculate the lower bound for a 90% confidence interval for the population mean cost of education. Round to the hundredth. c.) Calculate the upper bound for a 90% confidence interval for the population mean cost of...

Is the difference between the mean annual salaries of statisticians in Region 1 and Region 2...

Is the difference between the mean annual salaries of statisticians in Region 1 and Region 2 more than $8000? To decide, you select a random sample of statisticians from each region. The results of each survey are shown to the right. At α=0.10, what should you conclude? Answer A-E Region 1 x1 = $68,300 standard deviation = $8,750 n1 = 47 Region 2 x2 = $63,000 standard deviation = $9,275 n2 = 40 A. What are the alternative and null...