Question

In: Statistics and Probability

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.

 

 

Solutions

Expert Solution

Given that

n = 51, x̅ = 9.808 km        S.D = S = 5.013 km

d.f = n – 1 = 50

α = 1-95% = 1-0.96 = 0.05

tα/2 d.f = 2.01

 

Margin of error = critical value × s/√n

= 2.01 × 5.013/√51

M.E = 1.4099

 

∴ 95% confidence interval for population mean is

[x̅ ± M.E] = [9.808 ± 1.4099]

[8.3981, 11.2179]

 

∴95% confidence interval estimate of the mean epicenter depth of all earthquake lies between 8.3981 and 11.2179.


∴95% confidence interval estimate of the mean epicenter depth of all earthquake lies between 8.3981 and 11.2179.

Related Solutions

The random variable X has mean μ and standard deviation σ. A simple random sample of...
The random variable X has mean μ and standard deviation σ. A simple random sample of 50 values drawn from X has sample mean 14.7. Assuming the standard deviation is known to be 3.6, construct an 98% confidence interval for the value of μ. Put your answer in correct interval notation, (lower bound, upper bound), and round to three decimal places.
A simple random sample of size 44 has mean 3.01. The population standard deviation is 1.51....
A simple random sample of size 44 has mean 3.01. The population standard deviation is 1.51. Construct a 90% confidence interval for the population mean. 1.The parameter is the population (choose one) mean, standard deviation, proportion, variance 2. The correct method to find the confidence interval is the (choose one) z, t, chi square method
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.
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 sample of 51 babies, the mean weight is 21 pounds and the standard deviation...
In a sample of 51 babies, the mean weight is 21 pounds and the standard deviation is 4 pounds. Calculate a 95% confidence interval for the true mean weight of babies. Suppose we are interested in testing if the true mean weight of babies is 19.4 vs the alternative that it is not 19.4 with an alpha level of .05. Would this test be significant? Explain your answer. Perform the t test and use a t-table to get the p-value
A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation...
A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. *
Suppose x has a distribution with a mean of 70 and a standard deviation of 51....
Suppose x has a distribution with a mean of 70 and a standard deviation of 51. Random samples of size n = 36 are drawn. (a) Describe the  distribution. has a binomial distribution. has an approximately normal distribution.     has a geometric distribution. has a normal distribution. has an unknown distribution. has a Poisson distribution. (b) Compute the mean and standard deviation of the distribution. (For each answer, enter a number.) = = What price do farmers get for their watermelon crops?...
A simple random sample of 81 is selected from a population with a standard deviation of...
A simple random sample of 81 is selected from a population with a standard deviation of 17. The degree of confidence is 90%. What is the margin of error for the mean?
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...
Suppose ? is a random variable with mean ?=50 and standard deviation ?=49. A random sample...
Suppose ? is a random variable with mean ?=50 and standard deviation ?=49. A random sample of size ?=38 is selected from this population. a) Find ?(?¯<49).P(X¯<49). Round your answer to four decimal places. b) Find ?(X¯≥52).P(X¯≥52). Round your answer to four decimal places. c) Find ?(49.5≤X¯≤51.5).P(49.5≤X¯≤51.5). Give your answer to four decimal places. d) Find a value ?c such that ?(X¯>?)=0.15.P(X¯>c)=0.15. Round your answer to two decimal places.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT