Construct a 99% confidence interval for the population mean, based on the following sample data: 16.3, 15,9, 15.4, 16.7, and 16.9. Assume sampling from a normal population. Somple meen 1 16.3+ 15. 13+ 15.4 +16.7+16. = 0.49 2 5 16.04 5 Standard deviation. (16.3 - 16.04)² = 0.0672 (15.9-16.04)² 8 0186 ? 31-672 0.4356 (15.4-16.041 20.4086 2 (16.7-16-04)² = 116.8-16.0412 = 0.7396 82 1.1677 20.646

In a random sample of eight people, the mean driving distance to work was 24 2 miles and the standard deviation was 7.5 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean p. Interpret the results. Identify the margin of error. (Round to one decimal place as needed.)

Consider the following random sample from a normal population: 14,10,13,16,12,18,15, and 11. What is the 95% confidence interval for the population variance? A) 6.54 to 38.82 B) 11.39 to 15.86 C) 1.12 to 5.69 D) 3.11 to 29.51

A company that receives shipments of batteries tests a random sample of nine of them before agreeing to take a shipment. The company is concerned that the true mean lifetime for all batteries in the shipment should be at least 50 hours. From past experience it is safe to conclude that the population distribution of lifetimes is normal with a standard deviation of 3 hours. For one particular shipment the mean lifetime for a sample of nine batteries was 48.2 hours. a. Test, at the 10% level, the null hypothesis that the population mean lifetime is at least 50 hours. b. Find the power of a 10%-level test when the true mean lifetime of batteries is 49 hours.

10 squirrels were found to have an average weight of  610  grams with a sample standard deviation of  7.75 . Find the  95 % confidence interval of the true mean weight  (assume the t-student distribution). A.)(608.25, 614.16) B.)(561.02, 614.16) C.) (604.46,615.54) D.) (513.79, 612.77)

A simple random sample of 5 observations from a normal distribution was taken, and the following values were obtained: {15, 19, 19, 20, 22}. A 95% confidence interval for the mean is O A. (16.569, 21.431) O B. (15.835, 22.165) O C.(16.069, 21.931) O D. (17.124, 20.876) O E. (16.765, 21.235)

Consider the following random sample from a normal population: 14, 10, 13, 16, 12, 18, 15, and 11. WWhat is the 95% confidence interval for the population variance? O A. 6.54 to 38.82 O B. 1.12 to 5.69 O C. 11.39 to 15.86 O D. 3.11 to 29.51

53.305) B. (24.591, 35.563) C. (26.591, 37.563) D. (28.591, 39.563) 34. A simple random sample of size n = 36 is taken from a normally distributed original population. The sample standard deviation iss = 7. Find the 99% confidence interval for the population variance o2. A. (27.453, 98.756) B. (28.453, 99.756) C. (29.453, 100.756) D. (30.453, 101.756)

A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $19.26, $21.87, $15.83, $20.57, $19.31, $24.74, $22.82 Construct the 95% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Copy Step 2 of 4: Calculate the sample standard deviation for the given sample data. Round your answer to two decimal places.

The following are the normalized levels of a critical protein in 12 samples of blood: 8.2 , 4.7 , 10.3 , 11.7 , 18.3 , 5.9 , 3.8 , 18.9 , 7.6 , 14.2 , 9.8 , 16.4 Compute the sample mean, the sample median, and the sample standard deviation.