A random sample of n measurements was selected from a population with unknown mean u and known standard deviation. Calculate a 90% confidence interval for u for each of the situations given in parts a through e. a. n= 70, x= 26, o² = 13 ( 25.29 , 26.71 ) (Round to two decimal places as needed.) b. n= 150, x= 118, o² = 24 ( 117.34 , 118.66 ) (Round to two decimal places as needed.) c.n= 90, x= 18, o = 0.2 (Round to two decimal places as needed.)

The mean time between major failures of a group of milling machines is 4 years, with a standard deviation of 0.8. what is the minimum precent of milling machines that fail between 2 and 6 years show all work

The life of Sunshine CD players is normally distributed with a mean of 4.1 years and a standard deviation of 1.3 years. A CD player is guaranteed for 3 years. We are interested in the length of time a CD player lasts. X~

n= 50 Mean, = ΣΧ = 1570/50 = 31.4 Standard deviation, o = = 20.41 Using empirical rule, 68% of the data falls within one standard deviation, 95% within two standard deviations, and 99.7% within three standard deviations from the mean. => 68% of data falls between e + o = 31.4 + 20.41 = (10.99, 51.81) => 95% of the data falls between e ± 2 * o = 31.4 + 2*20.41 = (-9.42, 72.22) => 99.7% of the data falls between + 3 * o = 31.4 +3*20.41 = (-29.83, 92.63) QUESTIONS: 1. How does answers in 68%, 95%, and 99.7% of data compare, respectively? 2. How does it support the conclusion you reached in question 1? Explain thoroughly.

In a random sample of size 11, N(mu, phi) population, obtain 90% confidence interval for standard deviation (phi) if the sample variance s^2 = 28.

Assume the sample variances to be continuous measurements. Find the probability that a random sample of 30 observations, from a normal population with variance^2 = 8, will have S^2 greater than 10.78.

A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.07 years, with sample standard deviation s = 0.82 years. However, it is thought that the overall population mean age of coyotes is u = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use a = 0.01. (a) What is the level of significance? State the null and alternate hypotheses. о н: и 3 1.75 yг; H,: и # 1.75 yr O Ho: H > 1.75 yr; H,: µ = 1.75 yr Ο H9: μ 1.75 yr о н: и 3 1.75 yг; H,: и 0.250 O 0.100 < P-value < 0.250 O 0.050 < P-value < 0.100 0.010 < P-value < 0.050 O P-value < 0.010

The number of deaths per day in a city due to road accidents and due to other causes independently follows Poisson distribution with parameter 2 and 6 respectively. Find the probability that the total number of deaths on a particular day is 2 or fewer. (Given e^ -4 =0.018)

A random sample of 41 adult coyotes in a region of northern Minnesota showed the average age to be x = 2.09 years, with sample standard deviation s = 0.88 years. However, it is thought that the overall population mean age of coyotes is ? = 1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use ? = 0.01. (a) State the null and alternate hypotheses.(Select the correct one) H0: ? = 1.75 yr; H1: ? ≠ 1.75 yr H0: ? > 1.75 yr; H1: ? = 1.75 yr     H0: ? = 1.75 yr; H1: ? > 1.75 yr H0: ? = 1.75 yr; H1: ? < 1.75 yr (b) What is the value of the sample test statistic? Round your answer to three decimal places.     (c) Estimate the P-value.     (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ??(Selet which one is correct) We fail to reject the null hypothesis because the p-value is greater than ? = 0.01.…

A regression analysis yields the following information: Y^=2.21+1.49X ; n = 10; sy∙x = 1.66; ΣX = 32; ΣX2 = 134;  ∑(X−X¯¯¯¯)2=31.6 Compute the 95% confidence interval when X = 4.