Question

In: Statistics and Probability

hw18#4 A random sample of 80 observations produced a mean of x¯=21.1 from a population with...

hw18#4

A random sample of 80 observations produced a mean of x¯=21.1 from a population with a normal distribution and a standard deviation σ=4.54.

(a) Find a 90% confidence interval for ?μ
_____ ≤ ? ≤ ______

(b) Find a 99% confidence interval for ?μ
____ ≤ ? ≤ ______

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 21.1

Population standard deviation =    = 4.54

Sample size = n = 80

(a) At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z/2 * ( /n)

= 1.645 * ( 4.54 /  80 )

= 0.83

At 90% confidence interval estimate of the population mean is,

- E < < + E

21.1 - 0.83 <   < 21.2 + 0.83

20.27 <   < 21.93

( 20.27 , 21.93 )

The 90% confidence interval for is :  20.27 <   < 21.93

(b) At 99% confidence level

= 1 - 99%  

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z/2 * ( /n)

= 2.576 * ( 4.54 /  80 )

= 1.31

At 99% confidence interval estimate of the population mean is,

- E < < + E

21.1 - 1.31 <   < 21.2 + 1.31

19.79 <   < 22.41

( 19.79 , 22.41 )

The 99% confidence interval for is : 19.79 <   < 22.41


Related Solutions

A random sample of n = 50 observations from a quantitative population produced a mean x...
A random sample of n = 50 observations from a quantitative population produced a mean x = 2.3 and a standard deviation s = 0.34. Your research objective is to show that the population mean μ exceeds 2.2. Calculate the p-value for the test statistic z = 2.08. (Round your answer to four decimal places.) p-value =
A random sample of 100100 observations produced a mean of x¯¯¯=38.6x¯=38.6 from a population with a...
A random sample of 100100 observations produced a mean of x¯¯¯=38.6x¯=38.6 from a population with a normal distribution and a standard deviation σ=4.42σ=4.42. (a) Find a 9090% confidence interval for μμ ≤μ≤≤μ≤ (b) Find a 9595% confidence interval for μμ ≤μ≤≤μ≤ (c) Find a 9999% confidence interval for μμ ≤μ≤≤μ≤
A random sample of 120 observations produced a mean of ?⎯⎯⎯ =29.4 from a population with...
A random sample of 120 observations produced a mean of ?⎯⎯⎯ =29.4 from a population with a normal distribution and a standard deviation ?=2.45. (a) Find a 95% confidence interval for ? ___ ≤ ? ≤ ___ (b) Find a 90% confidence interval for ? ___ ≤ ? ≤ ___ (c) Find a 99% confidence interval for ?μ ___ ≤ ? ≤ ___
A random sample of 90 observations produced a mean of 32.4 from a population with a...
A random sample of 90 observations produced a mean of 32.4 from a population with a normal distribution and a standard deviation ?=2.98. (a) Find a 90% confidence interval for μ ≤?≤ (b) Find a 95% confidence interval for μ ≤?≤ (c) Find a 99% confidence interval for μ ≤?≤
A random sample of 100 observations from a quantitative population produced a sample mean of 29.8...
A random sample of 100 observations from a quantitative population produced a sample mean of 29.8 and a sample standard deviation of 7.2. Use the p-value approach to determine whether the population mean is different from 31. Explain your conclusions. (Use α = 0.05.) State the null and alternative hypotheses. (Choose Correct Letter) (a) H0: μ = 31 versus Ha: μ < 31 (b) H0: μ ≠ 31 versus Ha: μ = 31     (c) H0: μ < 31 versus Ha:...
A random sample of 130 observations produced a mean of ?⎯⎯⎯=36.1x¯=36.1 from a population with a...
A random sample of 130 observations produced a mean of ?⎯⎯⎯=36.1x¯=36.1 from a population with a normal distribution and a standard deviation σ=4.87. (a) Find a 95% confidence interval for μ ≤ μ ≤ (b) Find a 99% confidence interval for μ ≤ μ ≤ (c) Find a 90% confidence interval for μ ≤ μ ≤
A random sample of n = 1,400 observations from a binomial population produced x = 527...
A random sample of n = 1,400 observations from a binomial population produced x = 527 successes. You wish to show that p differs from 0.4. Calculate the appropriate test statistic. (Round your answer to two decimal places.) z = Calculate the p-value. (Round your answer to four decimal places.) p-value = Do the conclusions based on a fixed rejection region of |z| > 1.96 agree with those found using the p-value approach at α = 0.05? A.Yes, both approaches...
A random sample of n = 1,400 observations from a binomial population produced x = 667...
A random sample of n = 1,400 observations from a binomial population produced x = 667 successes. You wish to show that p differs from 0.5. Calculate the appropriate test statistic. (Round your answer to two decimal places.) z = Calculate the p-value. (Round your answer to four decimal places.) p-value =
A random sample of n = 500 observations from a binomial population produced x = 250...
A random sample of n = 500 observations from a binomial population produced x = 250 successes. Find a 90% confidence interval for p. (Round your answers to three decimal places.) Interpret the interval: A. In repeated sampling, 90% of all intervals constructed in this manner will enclose the population proportion. B. 90% of all values will fall within the interval. C. In repeated sampling, 10% of all intervals constructed in this manner will enclose the population proportion. D. There...
(4.1) A random sample of 89 observations produced a mean x = 26.1 and a standard...
(4.1) A random sample of 89 observations produced a mean x = 26.1 and a standard deviation s = 2.4 a. Find a? 95% confidence interval for ?. b. Find a? 90% confidence interval for ?. c. Find a? 99% confidence interval for ?. a. The? 95% confidence interval is (----,----) ?(Use integers or decimals for any numbers in the expression. Round to two decimal places as? needed.) b. The? 90% confidence interval is ----,------. (Use integers or decimals for...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT