Statistics 350 ? Fall 2009 ? HW6 Solutions 1. Walking a Straight Line (Part 1) ? In a past study 1 , a total of 20 randomly selected healthy men had their cadence measured (strides per second) during normal walking. a. A summer intern analyzes the data and reports the mean as 0.93 strides and the standard error of the mean (SEM) as 0.018 strides. Let the parameter ? be defined as the average cadence for all healthy men. Provide a 95% confidence interval estimate for ?. [1 point] Note: df = n ? 1 = 19 so t* = 2.09 x ± t* [ s.e.( x )] ? 0.93 ± 2.09*0.018 95% Confidence Interval: __0.89238__ to __0.96762__ b. Provide an interpretation of the 95% confidence interval found in part (a). [1 point] With 95% confidence, we estimate the true population mean cadence for all healthy men to be between 0.89238 and 0.96762 steps. c. Provide an interpretation of the 95% confidence level that was used to make the interval in part (a). [2 points] If the procedure is repeated many times, we would expect about 95% of the resulting confidence intervals to contain the true population mean cadence for all healthy men. d. The summer intern forgot to report the standard deviation for the cadences of these 20 men. Report that value. [1 point] nxess n s xes *).(.).(. =?= ? 0.018 *?(20) = 0.0805 Sample standard deviation s = _0.0805_ e. In addition to a requiring the data be a random sample (or a sample that is representative enough of cadences for the population of healthy men), there is an additional data assumption to consider. Brieftly state that condition and also how you would assess that condition. [2 points] Condition: The response (cadence measured as strides per second during normal walking) has a normal distribution for the population. How to assess that condition: This can be assessed by examining the data (the 20 observations) graphically: (1) Look at the histogram: does it appear to be approximately bell?shaped?; (2) Look at the qq plot: Do the points appear to fall approximately along a straight line with a positive slope? f. [0.5 points] Using just your common sense, had you calculated a 99% Confidence Interval, it would have been (circle one) narrower wider than the 95% Confidence Interval reported in part a. g. [0.5 points] To verify part (f), the t*multiplier in part a for a 95% Confidence Interval was _2.09__ as compared to a t* multiplier of __2.86__ that would have been used for the 99% Confidence Interval. 1 ?Can We Really Walk Straight??, Amer. J. of Physical Anthropology (1992): 19?27 Page 1 of 2 2. Walking A Straight Line (Part 2) ? Suppose the distribution of cadence for all healthy men is normally distributed with a mean of 0.90 strides and a standard deviation of 0.07 strides. a. Consider all such random samples of size n = 20, each sample providing a sample mean value x . What would be the standard deviation of those possible sample mean values? [1 point] Standard Deviation ( x ) = _0.01565_ b. Provide a detailed sketch of what the smoothed out histogram of those possible x values would look like. [2 points] 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.05 1.10 x values By Empirical Rule, about 99.7% of the sample means ( values) will be between 0.853 and 0.947. c. What is the probability that of seeing a sample mean as small as 0.88 strides or even less? [2 points] probability: ____0.1003 or 10.3%_____ d. Suppose the distribution of cadence had not been normally distributed, but rather highly skewed to the right. Would you have been able to answer the probability in part (c) as you did? [1 point] Circle one: Yes No Explain: No, we could not apply the same technique because we do not have a large enough sample size (e.g. n > 25) to rely on the Central Limit Theorem. e. Suppose the distribution of cadence had been highly skewed to the right and the sample size had been n = 100. Give the name of the result that would have allowed you to still answer the probability question in part (c). [1 point] Now the sample size is large enough (n > 25) to apply the Central Limit Theorem. By the Central Limit Theorem, we can reasonably approximate the distribution of with a normal model. Page 2 of 2 Brenda Gunderson 350 HW 1
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