A sample of households in Colorado shows the following statistics about household income:
Mean: 65000
Standard Deviation: 5500
Sample Size: 15
Confidence Level: 85 %
What is the t-value you should use to create a confidence interval in this situation? Submit the absolute value rounded to 2 decimal places.
A sample of households in Colorado shows the following statistics about household income: Mean: 65000 Standard...
24.The mean monthly expenditure on gasoline per household in Middletown is determined by selecting a random sample of 100 households. The sample mean is $115, with a sample standard deviation of $38. What is the upper bound of a 80% confidence interval for the mean monthly expenditure on gasoline per household in Middletown?$118.42$116.38$121.66- $149.20$119.8725.You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample mean is $ 800...
Use the one-mean t-interval procedure with the sample mean, sample size, sample standard deviation, and confidence level given below to find a confidence interval for the mean of the population from which the sample was drawn. x̄=4.0 n=61 s=6.1 confidence level =99% The 99% confidence interval about μ is ??? to ??? (Round to four decimal places as needed.)
25> Consider a variable known to be Normally distributed with unknown mean μ and known standard deviation σ-10. (a) what would be the margin of error of a 95% confidence interval for the population mean based on a random sample size of 25? The multiplier for a z confidence interval with a 95% confidence level is the critical value z. 1.960. (Enter your answer rounded to three decimal places.) margin of error 25 (b) What would be the margin of...
How many computers? In a simple random sample of 185 households, the sample mean number of personal computer ls was 1.64. assume the population of the standard deviation is 0.47 a) construct a 99.9% confidence interval for the mean number of personal computers. round the answer to at least two decimal places. a 99.9% confidence interval for the mean number of personal computers is ---< mean <----- 2) In a sample of 75 clamps, the mean step to complete this...
A sample of 10 students record their scores on the final exam for their statistics class. The mean of the sample is 81 with sample standard deviation 7 points. Analysis of the 10 sample values indicated that the population is approximately normal. We wish to find the 95% confidence interval for the population mean test scores. What is the confidence level, c? Which of the following is correct? To find the confidence interval, a z-critical value should be used because...
A
random sample of 15 statistics textbooks has a mean price of $105
with a standard deviation of $30.25. Assume the distribution of
statistics textbook prices is not normally distributed.
10:13 Notes Done (c) A random sample of 15 statistics textbooks has a mean price of $105 with a standard deviation of $30.25. Assume the distribution of statistics textbook prices is not normally distributed. i) Critical value at 85% Confidence Level (indicate za/2 or ta/2): ii) Critical value at 95%...
Use the one-mean t-interval procedure with the sample mean, sample size, sample standard deviation, and confidence level given below to find a confidence interval for the mean of the population from which the sample was drawn. sample mean=3.0 n=41 s=5.4 confidence level=90% The 90% confidence interval about μ is ?? to ???
A sample mean, sample size, population standard deviation, and confidence level are provided. Use this information to complete parts (a) through (c) x = 33, n = 25, C = 6, confidence level = 90% Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table a. Use the one-mean z-interval procedure to find a confidence interval for the mean of the population from which the sample...
A simple random sample of 10 households, the number of TV's that each household had is as follows: 4 , 2 , 3 , 1 , 2 , 1 , 1 , 4 , 1 , 2 Assume that it is reasonable to believe that the population is approximately normal and the population standard deviation is 0.49 . What is the lower bound of the 95% confidence interval for the mean number of TV's?
Here is an example with steps you can follow: sample size n=9, sample mean=80, sample standard deviation s=25 (population standard deviation is not known) Estimate confidence interval for population mean with confidence level 90%. Confidence Interval = Sample Mean ± Margin of Error Margin of Error = (t-value)×s/√n t-value should be taken from Appendix Table IV. For n=9 df=n-1=9-1=8 For Confidence Level 90% a = 1 - 0.90 = 0.10, a/2 = 0.10/2 = 0.05 So, we are looking for...