Question

In: Statistics and Probability

The following summary statistics were derived from a simple random sample: Mean=63.7 Standard deviation=14.1 n=51 Compute...

The following summary statistics were derived from a simple random sample:

Mean=63.7

Standard deviation=14.1

n=51

Compute a 99% confidence interval for the population mean. Report the upper limit for your answer.

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 63.7

sample standard deviation = s = 14.1

sample size = n = 51

Degrees of freedom = df = n - 1 = 51 - 1 = 50

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

t,df = t0.01,50 = 2.403

Margin of error = E = t,df * (s /n)

= 2.403 * (14.1 / 51)

Margin of error = E = 4.7

At 99% upper confidance level is

+ E

63.7 + 4.7

68.4

The upper limit : 68.4

orchestra answered 2 years ago
Next > < Previous

Related Solutions

A statistics practitioner took a random sample of 51 observations from a population whose standard deviation...
A statistics practitioner took a random sample of 51 observations from a population whose standard deviation is 28 and computed the sample mean to be 108. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99%...
A statistics practitioner took a random sample of 51 observations from a population whose standard deviation...
A statistics practitioner took a random sample of 51 observations from a population whose standard deviation is 30 and computed the sample mean to be 99. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99%...
A simple random sample with n = 56 provided a sample mean of 26.5 and a sample standard deviation of 4.4.
  A simple random sample with n = 56 provided a sample mean of 26.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.) (a) Develop a 90% confidence interval for the population mean. to (b) Develop a 95% confidence interval for the population mean. to (c) Develop a 99% confidence interval for the population mean. to (d) What happens to the margin of error and the confidence interval as the confidence level is increased?...
In a simple random sample of 51 community college statistics students, the mean number of college...
In a simple random sample of 51 community college statistics students, the mean number of college credits completed was x=50.2 with standard deviation s = 8.3. Construct a 98% confidence interval for the mean number of college credits completed by community college statistics students. Verify that conditions have been satisfied: Simple random sample?                            n > 30? Find the appropriate critical value Since σ is unknown, use the student t distribution to find the critical value tα2 on table A-3 Degrees...
A simple random sample of epicenter depths of 51 earthquakes has a mean of 9.808 kilometers (km) and a standard deviation of 5.013 km
A simple random sample of epicenter depths of 51 earthquakes has a mean of 9.808 kilometers (km) and a standard deviation of 5.013 km. Determine the critical value of t and the margin of error, and then construct the 95% confidence interval estimate of the mean epicenter depth of all earthquakes.  
A simple random sample with n=54   provided a sample mean of 21.0 and a sample standard...
A simple random sample with n=54   provided a sample mean of 21.0 and a sample standard deviation of 4.7 . a. Develop a 90% confidence interval for the population mean (to 1 decimal). , b. Develop a 95% confidence interval for the population mean (to 1 decimal). , c. Develop a 99% confidence interval for the population mean (to 1 decimal). , d. What happens to the margin of error and the confidence interval as the confidence level is increased?...
A simple random sample with n=50 provided a sample mean of 2.3 and a sample standard...
A simple random sample with n=50 provided a sample mean of 2.3 and a sample standard deviation of 4.7. a. Develop a 90% confidence interval for the population mean (to 2 decimals). , b. Develop a 95% confidence interval for the population mean (to 2 decimals). , c. Develop a 99% confidence interval for the population mean (to 2 decimals). ,
A simple random sample with n=56 provided a sample mean of 24 and a sample standard...
A simple random sample with n=56 provided a sample mean of 24 and a sample standard deviation of 4.6. a. Develop a 90% confidence interval for the population mean (to 1 decimal). , b. Develop a 95%confidence interval for the population mean (to 1 decimal). , c. Develop a 99% confidence interval for the population mean (to 1 decimal). , d. What happens to the margin of error and the confidence interval as the confidence level is increased?
A simple random sample with n=54 provided a sample mean of 22.5 and a sample standard...
A simple random sample with n=54 provided a sample mean of 22.5 and a sample standard deviation of 4.2. a. Develop a 90% confidence interval for the population mean (to 1 decimal). ( ______ , ______ ) b. Develop a 95% confidence interval for the population mean (to 1 decimal). ( ______ , _______ ) c. Develop a 99% confidence interval for the population mean (to 1 decimal). ( ______ , _______ )   ,   d. What happens to the margin...
A simple random sample with n=56 provided a sample mean of 23.5   and a sample standard...
A simple random sample with n=56 provided a sample mean of 23.5   and a sample standard deviation of 4.2 Develop a 90 % confidence interval for the population mean (to 1 decimal). Develop a 95 % confidence interval for the population mean (to 1 decimal) Develop a   99 % confidence interval for the population mean (to 1 decimal).
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT