Patients with an average blood pressure of 140 and above are classified as having hypertension. Over the course of 8 visits, the patient's blood pressure is measured as 137 143 141 133 We would like to rule out the possibility that the patient has hypertension, so we put the data into statistical analysis software: N 8 144 130 131 140 Reading the output, we find that Mean 137.4 • The sample size is n = • The sample mean is = • The sample standard deviation is SD = . From this we can calculate the standard error to SE • Calculate the t-score t = x 140 SE • The degrees of freedom for the t-score is n-1= SD √n Note that the observed sample mean is is less than than 140. This could mean that the patient does not have hypertension, or they could have hypertension, but the small numbers were due to sampling variability. To see if this is the case, we will test the hypothesis that μ = 140 against the alternative that u 140 at a significance level of a = 0.05. = Std. Dev. 5.5

Consider the given data set. n = 12 measurements: 9, 8, 1, 5, 5, 7, 5, 6, 8, 7, 2, 0 Find the mean. USE SALT Find the standard deviation. (Round your answer to four decimal places.) z = Find the z-score corresponding to the minimum in the data set. (Round your answer to two decimal places.) Find the z-score corresponding to the maximum in the data set. (Round your answer to two decimal places.) z = Do the z-scores indicate that there are possible outliers in the data set? Since the z-score for the larger observation is larger than 2 in absolute value, the larger value is unusually large. Since both z-scores exceed 2 in absolute value, both of the observations are unusual. Since neither z-score exceeds 2 in absolute value, none of the observations are unusually small or large. Since the z-score for the smaller observation is larger than 2 in absolute value, the smaller value is unusually small.

Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What can be concluded? 143 140 141 136 133 Right arm Left arm 180 174 192 140 144 In this example, . is the mean value of the differences d for the population of all pairs of data, where each individual difference d is defined as the measurement from the right arm minus the measurement from the left arm. What are the null and alternative hypotheses for the hypothesis test? O A. Ho: Ha = 0 O B. Ho: Hd #0 0 = Prt :H O D. Ho: Hd =0 H O C. Ho: Ha 0 Identify the test statistic. t%3D (Round to two decimal places as needed.) Identify the P-value. P-value (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test?…

Consider the following data: −10, −10, 0, 3, −10, 3, 3  Step 1 of 3: Calculate the value of the sample Variance. Round your answer to one decimal place. Step 2 of 3: Calculate the value of the sample deviation. Round your answer to one decimal place. Step 3 of 3: Calculate the value of the range.

What was the age distribution of prehistoric Native Americans? Extensive anthropological studies in the southwestern United States gave the following information about a prehistoric extended family group of 84 members on what is now a Native American reservation. For this community, estimate the mean age expressed in years, the sample variance, and the sample standard deviation. For the ciass 31 and over, use 35.5 as the class midpoint. (Round your answers to one decimal place.) Age range (years) Number of individuals 1-10 11-20 19 21-30 15 31 and over 40 10