Note: There are three possible conclusions that can be inferred from a confidence interval involving the difference of two dependent sample means. * If the ENTIRE interval is positive, then Ma > 0 and, therefore, M1 > Hz. * If the ENTIRE interval is negative, then Ma TESTS -> 8:TInterval (because the critical value & test statistic are t-scores, and we are doing a confidence interval.). Use the DATA option as you have the raw data. Remember you are using the difference data in L3. Input the required information. Write what you see below.Confidence Intervals Involving the Means of Two Dependent Samples (Matched Pairs) A drug company is analyzing the effectiveness of a blood pressure reducing medication. A random sample of 9 subjects with high blood pressure were tested before and after three weeks normal. of treatment with the drug and data appear in the table below. Assume blood pressure levels are Patient m + in Level before 170 8 9 179 182 193 209 185 169 Level after 155 210 167 159 151 176 183 159 145 146 177 Find (& interpret) a 90% confidence interval for the mean difference in blood pressure level after treatment . a. Requirement Check (Remember to make sure BOTH samples meet each requirement!) b. Write down known information and hypotheses. Population # 1 = Population # 2 = d = , Sd = , n = , a =, Calculate the margin of error. d. f . = to E = tc . -a d. Make the confidence interval d - E Calc -> 1-Var Stats on L3 to get d (x of the difference lists) and Sa (standard deviation of the difference list). Look at the x line to find d = Look at the sx line to find Sa = Our null hypothesis for paired data will always be that the mean difference (Ma) between thee. Compare the p-value to a, and write your conclusion in real world context. f. Can we use the regression equation to predict the cost of a Miata that has been driven 20,000 miles? If so what is the cost? If not, why not?1. Prof. Bodrero wants to buy a convertible and is researching prices of used Miatas. A random sample of 7 cars found the following information on number of miles driven and cost of the car. Assume that for a fixed # of miles, cost is normally distributed and for a fixed cost, # miles driven is normally distributed. Using a = 5%, test for linear correlation between # miles driven and cost of the car. Source: kbb.com. # miles 2728 1037 11625 35925 10644 Cost, $ 28442 30456 13991 26828 23323 21900 17895 12000 18995 a. Requirement Check b. Write down the known information and hypotheses n = a = (Recall that p represents the population correlation coefficient.) Ho: p = 0 (which means H1: p 0 (which means > 0 indicates a positive correlation. TEST -> F:LinRegTTest. Identify the XList as L1 and the YList as L2. Then choose the appropriate alternative hypothesis. p-value:pairs of measurements is equal to 0, denoted by / d = 0. As usual, the alternate hypothesis will take one of three forms Ma 0, or Ha # 0, depending on the problem. Ho: Hd = 0 H1: Ha _0 (, based on the problem) Type of test: - tailed a = . Find the critical values. Then draw and label the picture of the distribution. d. f . = n - 1 = to = - 1 0 d. Find the test statistic t = 3d Vn Then label the test statistic on the picture. e. Use the test statistic to find the p-value f. Using the test statistic and the p-value, state your conclusion. g. Now let's use your TI-83/84 calculator to find the test statistic and p-value Go to STAT -> TEST -> 1: T-Test (because the critical value & test statistic are t-scores, and we are doing a hypothesis test.). Use the DATA option as you have the raw data. Remember you are using the difference data in L3. Input the required information Write down what you see in the space below