Question 6 (1 point) You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 10 customers and find that the average dollar amount spent per transaction per customer is $83.652 with a standard deviation of $15.2173. Create a 99% confidence interval for the true average spent for all customers per transaction. 1) (-68.013, 99.291 ) 2) (68.013.99.291) 3) (68.402.98.902) 4) (78.84, 88.464) 5) (80.402, 86.902) Question 9 (1 point) A new drug to treat high cholesterol is being tested by pharmaceutical company. The cholesterol levels for 12 patients were recorded before administering the drug and after. The mean difference in total cholesterol levels (after-before) was 15.069 mg/dL with a standard deviation of 18.4384 mg/dL. Create a 99% confidence interval for the true average difference in cholesterol levels by the drug. 1) (-1.4623, 31.6003) 2) (-1.1894, 31.3274) 3) (1.4623, 31.6003) 4) (9.7463, 20.3917) 5) (11.9632, 18.1748)

Question 6: Correct answer: Option (2) (68.013,99.291) since Question 9: Correct answer: Option (1) (-1.4623, 31.6003) since The provided sample mean is X = 83.652 and the sample standard deviation is 8 = 15.2173. The size of the sample is n = 10 and the required confidence level is 99% The number of degrees of freedom are df = 10-1=9, and the significance level is a = 0.01. Based on the provided information, the critical t-value for a = 0.01 and df = 9 degrees of freedom is te = 3.25.The 99% confidence for the population mean u is computed using the following expression C1 = (x ) Therefore, based on the information provided, the 99 % confidence for the population mean u is CI = 83.652 3.25 x 15.2173 10 .83.652 + .05.052 + 10 ) = (83.652 - 15.639,83.652 +15.639) = (68.013, 99.291) The provided sample mean is X = 15.069 and the sample standard deviation is 8 = 18.4384. The size of the sample is n = 12 and the required confidence level is 99% The number of degrees of freedom are df = 12-1= 11, and the significance level is a = 0.01 Based on the provided information, the critical t-value for a = 0.01 and df = 11 degrees of freedom is te = 3.106. The 99% confidence for the population mean y is computed using the following expression CI = (* * * * * x + kovin) Therefore, based on the information provided, the 99% confidence for the population mean u is CI = ( 15.069 - Geo 3.106 x 18.4384 0 712 3.106 18.4384 12 ) = (15.069 - 16.531, 15.069 +16.531) = (-1.462, 31.6) .

Question 6 (1 point) You own a small storefront retai…

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The provided sample mean is X = 83.652 and the sample standard deviation is 8 = 15.2173. The size of the sample is n = 10 and

The provided sample mean is X = 15.069 and the sample standard deviation is 8 = 18.4384. The size of the sample is n = 12 and

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