10.2 - Key

) 2 + = ( 5.5 0.233 ) 1.86 · ( 2.517 ) 2 ( 3.292 ) 2 + n1 12 9 = ( 2.698 , 7.836 ) We are 90 % confident that the interval from 2.698 to 7.836 contains the true mean difference in polyphenol levels for healthy men who drink half a bottle of white wine a day and healthy men who drink half a bottle of red wine a day . This provides evidence that there is in fact a difference , since the interval does not contain zero . College financial aid offices expect students to use summer earnings to help pay for college . But how large are these earnings ? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned . The financial aid office separated the responses into two groups based on gender . Here are the data in summary form : Grou p n Males 675 $ 1884.52 Female s 621 S x $ 1368.3 7 $ 1360.39 $ 1037.46 How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right ? Since income is not something that can be negative and two standard deviations below the mean would be a large negative number , the data cannot be normal . Therefore , they must be skewed right . The use of two - sample t procedures is still justified . Why ? Since both sample sizes are greater than 30 , we can still use two - sample t procedures because the CLT allows us to assume the sampling distribution will be approximately normal Construct and interpret a 90 % confidence interval for the difference between the mean summer earnings of male and female students at this university . Be sure to include all four steps ( State , Plan , Do , Conclude)

upper : 0.248 df : 55 P - value : 0.4032 0.806 ( multiply by 2 for two tailed test ) Since the p - value is larger than the significance level of 0.05 we fail to reject the null hypothesis . There is not convincing evidence that there is a difference in the mean words per day spoken by women and men . In a pilot study , a company's new cholesterol - reducing drug outperforms the currently available drug . If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood ( mg / dl ) greater than with the current drug , the company will begin the expensive process of mass - producing the new drug . For the 14 subjects who were assigned at random to the current drug , the mean cholesterol reduction was 54.1 mg / dl with a standard deviation of 11.93 mg / dl . For the 15 subjects who were randomly assigned to the new drug , the mean cholesterol reduction was 68.7 mg / dl with a standard deviation of 13.3 mg / dl . Graphs of the data reveal no outliers or strong skewness . Carry out an appropriate significance test . What conclusion would you draw ? ( Note that the null hypothesis is not equal to zero ) μ ] = true mean mg / dl cholesterol reduction for patients on the new drug μ2 = true mean mg / dl cholesterol reduction for patients on the current drug Ho H1 H2 = 10 Ha 1-2 10 The subjects were randomly assigned to treatments , we were not sampling at random from a population so we don't need to check the 10 % condition . While both samples are less than 30 in size , the data did not show outliers or strong skewness , so all conditions for inference are met . Two sample t test with α = 0.05 x1 = 68.7 8x1 = 13.3 X254.1 t = tcdf = 5x2 = 11.93 (≈1- X2 ) — ( μ1μ2 ) ( -1 ) 2 ( 6x2 ) 2 M1 lower : 0.982 + ng