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ILI Was $1500 with a standard June the rent is normally distributed. Normal distribution B) -distribution...
8. Find the critical t values for the following: (Section 6.2) • C=0.99 and n =10 C=0.95 and n =16 • c=0.80 and n =250 c=0.98 and n =95 Note: If you cannot find the degrees of freedom (df) on the t-distribution chart, use the lower df.
Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can In a recent season, the population standard deviation of the yards per carry for all running backs was 1.21. The yards per carry of 25 randomly selected running backs 2.8 be used, explain why. Interpret the results are shown below. Assume the yards per carry are normally distributed 2.7 3.7 4.7 7.1 3.7 5.9 2.5...
6.2.36-1 Question Help Use the standard normal distribution or the t-distribution to construct a 95% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a random sample of 15 mortgage institutions, the mean interest rate was 3.42% and the standard deviation was 0.44%. Assume the interest rates are normally distributed Which distribution should be used to construct the confidence interval? O O O O O A. Use a...
7. Determine whether a normal distribution or a normal distribution or t distribution should be used or whether neither of should be used to construct a confidence interval for the mean. (Section 6.2) Circle the correct answer. in a random sample of 22 families, the mean weekly food expense was $95.60 with a sample standard deviation of $22.50. Assume the weekly food expense is normally distributed. A) Normal distribution B) t-distribution C) Neither For a sample of 20 IQ scores...
(a) find the margin of error for the values of c, s, and n, and (b) construct the confidence interval for μ using the t-distribution. Assume the population is normally distributed. 1. c=0.90, s=25.6, n=16, x= 72.1 2. c=0.95, s=1.1, n=25, x = 3.5 3. c = 0.98, s=0.9, n=12, x= 6.8 4. c = 0.99, s=16.5, n=20, x= 25.2
Use the standard normal distribution or the t-distribution to construct a 90% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a random sample of 13 people, the mean length of stay at a hospital was 6.9 days. Assume the population standard deviation is 2.1 days and the lengths of stay are normally distributed. Which distribution should be used to construct the confidence interval? A. Use a normal...
9. The lifetime of a certain brand of lightbulbs are assumed to be normally distributed. A researcher took a random sample of these lightbulbs and observed the following failure times: 17 | 17 | 24 24 | 25 34 40 Let xay be the critical value of a t random variable with v degrees of freedom. The following table lists values of x2., for specific combinations of a and v: v=6 v=7) a= 0.975 1.237 1.690 a = 0.95 1.635...
9. The lifetime of a certain brand of lightbulbs are assumed to be normally distributed. A researcher took a random sample of these lightbulbs and observed the following failure times 24 25 17 24 34 40 17 Let xbe the critical value of a t random variable with v degrees of freedom. The following table lists values of x for specific combinations of a and v a 0.05 Q=0.975 a=0.025 O= 0.95 1.237 14.449 v6 1.635 12.592 16.013 1.690 1.2.167...
Use the standard normal distribution or the t distribution to construct a 9 % confidence interval for the population mean Justify your decision if neither distribution can be used, explain why Interpret the results ln a random sample of 17 mortgage institutions, the mean interest rate was 3.69% and the standard deviation was 36% Assume the iterest rates are normally distributed Which distribution should be used to construct the confidence interval? ○ A. Use a t-distribution because it is a...
Use the standard normal distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a recent season, the population standard deviation of the yards per carry for all running backs was 1.27. The yards per carry of 25 randomly selected running backs are shown below. Assume the yards per carry are normally distributed. 2.9 8.4 7.2 4.3 6.8 2.7 7.2 4.8...