30 people are selected randomly from a certain town. If their mean age is 60.5 and ? = 4.6, find a 95% confidence interval for the true mean age, ?, of everyone in the town.

a confidence interval for a population mean has a margin of error of 0.081

A random sample of 100 light bulbs has a mean lifetime of 3000 hours. Assume that the population standard deviation of the lifetime is 500 hours. Construct a 95% confidence interval estimate of the mean lifetime. [B]2902 < u < 3098[/B] using our [URL='http://www.mathcelebrity.com/normconf.php?n=100&xbar=3000&stdev=500&conf=95&rdig=4&pl=Large+Sample']confidence interval for the mean calculator[/URL]

A random sample of 144 with a mean of 100 and a standard deviation of 70 is known from a population of 1,000. What is the 95% confidence interval for the unknown population? [URL='http://www.mathcelebrity.com/normconf.php?n=144&xbar=100&stdev=70&conf=95&rdig=4&pl=Large+Sample']Large Sample Confidence Interval Mean Test[/URL] [B]88.5667 < u < 111.4333[/B]

A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places) Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get [B]58.89 < u < 63.11[/B]

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.44 hours, with a standard deviation of 1.74 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesis (?1 - ?2). Using our confidence interval for [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=+2.31&n2=+40&xbar2=+4.44&stdev2=1.74&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means calculator[/URL], we get: [B]0.0278 < ?1 - ?2 < 1.5322[/B]

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.29 hours, with a standard deviation of 1.58 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1 - u2) What is the interpretation of this confidence interval? A. There is 90% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours B. There is a 90% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours C. There is 90% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours D. There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours 0.2021 < u1 - u2 < 1.6579 using our [URL='http://www.mathcelebrity.com/meandiffconf.php?n1=+40&xbar1=+5.22&stdev1=2.31&n2=40&xbar2=4.29&stdev2=1.58&conf=+90&pl=Mean+Diff+Conf.+Interval+%28Large+Sample%29']difference of means confidence interval calculator[/URL] [B]Choice D There is a 90% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours[/B]

A random sample of n = 10 flash light batteries with a mean operating life X=5 hr. And a sample standard deviation S = 1 hr. is picked from a production line known to produce batteries with normally distributed operating lives. What's the 98% confidence interval for the unknown mean of the working life of the entire population of batteries? [URL='http://www.mathcelebrity.com/normconf.php?n=10&xbar=5&stdev=1&conf=98&rdig=4&pl=Small+Sample']Small Sample Confidence Interval for the Mean test[/URL] [B]4.1078 < u < 5.8922[/B]

A survey was conducted that asked 1007 people how many books they had read in the past year. Results indicated that x overbarequals11.3 books and sequals16.6 books. Construct a 90% confidence interval for the mean number of books people read. Interpret the interval. x bar = 11.3 s = 16.6 n = 1007 [URL='https://www.mathcelebrity.com/normconf.php?n=1007&xbar=11.3&stdev=16.6&conf=90&rdig=4&pl=Not+Sure']We use our confidence interval calculator[/URL] and get [B]10.4395 < u < 12.1605[/B]. [B][I]We interpret this as: If we repeated experiments, the proportion of such intervals containing u would be 90%[/I][/B]

A teacher assumed that the average of grades for a math test was 80. Imagine 20 students took the test and the 95% confidence interval of grades was (83, 90). Can you reject the teacher's assumption? a. Yes b. No c. We cannot tell from the given information [B]a. Yes[/B] [I]At the 0.05 significance level, yes since 80 is not in the confidence interval.[/I]

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

Free Confidence Interval for the Mean Calculator - Calculates a (90% - 99%) estimation of confidence interval for the mean given a small sample size using the student-t method with (n - 1) degrees of freedom or a large sample size using the normal distribution Z-score (z value) method including Standard Error of the Mean. confidence interval of the mean

Free Confidence Interval for Variance and Standard Deviation Calculator - Calculates a (95% - 99%) estimation of confidence interval for the standard deviation or variance using the χ² method with (n - 1) degrees of freedom.

Free Confidence Interval of a Proportion Calculator - Given N, n, and a confidence percentage, this will calculate the estimation of confidence interval for the population proportion π including the margin of error. confidence interval of the population proportion

Free Confidence Interval/Hypothesis Testing for the Difference of Means Calculator - Given two large or two small distriutions, this will determine a (90-99)% estimation of confidence interval for the difference of means for small or large sample populations.
Also performs hypothesis testing including standard error calculation.

Construct a confidence interval of the population proportion at the given level of confidence. x = 120, n = 300, 99% confidence Round to 3 decimal places as needed [B]0.327 < p < 0.473[/B] using our [URL='http://www.mathcelebrity.com/propconf.php?bign=300&smalln=120&conf=99&pl=Proportion+Confidence+Interval']proportion confidence interval calculator[/URL]

Free Covariance and Correlation coefficient (r) and Least Squares Method and Exponential Fit Calculator - Given two distributions X and Y, this calculates the following:
* Covariance of X and Y denoted Cov(X,Y)
* The correlation coefficient r.
* Using the least squares method, this shows the least squares regression line (Linear Fit) and Confidence Intervals of α and Β (90% - 99%)
Exponential Fit
* Coefficient of Determination r squared r²
* Spearmans rank correlation coefficient
* Wilcoxon Signed Rank test

The data below consists of the pulse rates (in beats per minute) of 32 students. Assuming ? = 10.66, obtain a 95.44% confidence interval for the population mean. 80 74 61 93 69 74 80 64 51 60 66 87 72 77 84 96 60 67 71 79 89 75 66 70 57 76 71 92 73 72 68 74

Express the confidence interval 0.039 < p < 0.479 in the form of p ± E. We find the range of this interval: Range = Upper Bound - Lower Bound Range = 0.479 - 0.039 Range = 0.44 Each piece on opposite sides of p gets: 0.44/2 = 0.22 So our expression becomes [B]p ± 0.22 [MEDIA=youtube]FGZcvcuWCpE[/MEDIA][/B]

The data below consists of the pulse rates (in beats per minute) of 32 students. Assuming ? = 10.66, obtain a 95.44% confidence interval for the population mean. 80 74 61 93 69 74 80 64 51 60 66 87 72 77 84 96 60 67 71 79 89 75 66 70 57 76 71 92 73 72 68 74

Physiologists often use the forced vital capacity as a way to assess a person's ability to move air in and out of their lungs. A researcher wishes to estimate the forced vital capacity of people suffering from asthma. A random sample of 15 asthmatics yields the following data on forced vital capacity, in liters. 5.1 4.9 4.7 3.1 4.3 3.7 3.7 4.3 3.5 5.2 3.2 3.5 4.8 4.0 5.1 Use the data to obtain a 95.44% confidence interval for the mean forced vital capacity for all asthmatics. Assume that ? = 0.7.

Free Margin of Error from Confidence Interval Calculator - Given a confidence interval, this determines the margin of error and sample mean.

Free P-Hat Confidence Interval Calculator - Given a large sized distribution, and a success amount for a certain criteria x, and a confidence percentage, this will calculate the confidence interval for that criteria.

Free Paired Means Difference Calculator - Calculates an estimation of confidence interval for a small or large sample difference of data. Confidence interval for paired means

based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error

a confidence interval for a population mean has a margin of error of 0.081. Determine the length of the confidence interval

The NJ state education department finds that in a random sample of 100 persons who attended college, 40 received a college degree. What's the 95% confidence interval for the proportion of college graduates out of all the persons who attended college? [URL='http://www.mathcelebrity.com/propconf.php?bign=100&smalln=40&conf=95&pl=Proportion+Confidence+Interval']Proportion Confidence Interval Test[/URL] 0.304 < p < 0.496 --> [B]30.4% < p < 49.6%[/B]

The principal randomly selected six students to take an aptitude test. Their scores were: 87.4 86.9 89.9 78.3 75.1 70.6 Determine a 90% confidence interval for the mean score for all students.

First, determine the [URL='http://www.mathcelebrity.com/statbasic.php?num1=87.4%2C86.9%2C89.9%2C78.3%2C75.1%2C70.6&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']mean and standard deviation[/URL] for the [I]sample[/I] Mean = 81.3667 SD = 7.803 Next, use our [URL='http://www.mathcelebrity.com/normconf.php?n=6&xbar=81.3667&stdev=7.803&conf=90&rdig=4&pl=Small+Sample']confidence interval for the mean calculator[/URL] with these values and n = 6 [B]74.9478 < u < 87.7856[/B]

Which of the following descriptions of confidence interval is correct? (Select all that apply) a. If a 99% confidence interval contains 0, then the 95% confidence interval contains 0 b. If a 95% confidence interval contains 0, then the 99% confidence interval contains 0 c. If a 99% confidence interval contains 1, then the 95% confidence interval contains 1 d. If a 95% confidence interval contains 1, then the 99% confidence interval contains 1 [B]a. If a 99% confidence interval contains 0, then the 95% confidence interval contains 0 c. If a 99% confidence interval contains 1, then the 95% confidence interval contains 1 [/B] [I]The lower the confidence interval, the wider the range, so if a higher confidence interval contains a point, a lower confidence interval will contain that point as well.[/I]