Chapter 10 Solutions - 223 Chapter 10 10.1(a The sampling distribution of x is approximately normal with mean x = 280 and 60 = 2.0702(b The mean is 280

Chapter 10 Solutions - 223 Chapter 10 10.1(a The sampling...

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Chapter 10 10.1 (a) The sampling distribution of xis approximately normal with mean 280xµ=and standard deviation 602.0702840xnσσ==±. (b) The mean is 280. One standard deviation from the mean: 277.9 and 282.1; two standard deviations from the mean: 275.8 and 284.2; and three standard deviations from the mean: 273.7 and 286.3. Two different sketches are provided below. Sample meanNormal density curve286.3284.2282.1280.0277.9275.8273.70.200.150.100.050.00(c) 2 standard deviations; . (d) The confidence intervals drawn will vary, but they should all have length. See the sketch above on the right. (e) About 95% (by the 68-95-99.7 rule). 4.2m±28.4m±10.2 (a) Incorrect; the probability is either 0 or 1, but we don’t know which. (b) Incorrect; the general form of these confidence intervals is xm±, so xwill always be in the center of the confidence interval. (c) Incorrect; the different samples will yield different sample means, and the distribution of those sample means is used to provide an interval that captures the population mean. (d) Incorrect; there is nothing magic about the interval from this one sample. Our method for computing confidence intervals is based on capturing the mean of the population, not a particular interval from one sample. (e) Correct interpretation. 10.3 No. The student is misinterpreting 95% confidence. This is a statement about the mean score for all young men, not about individual scores. 10.4 (a) The sampling distribution of xis approximately normal with mean xµµ=and standard deviation 0.40.056650xnσσ==±. (b) The mean isµ. One standard deviation from the mean: 0.0566µand0.0566µ+; two standard deviations from the mean: 0.1132µand0.1132µ+; and three standard deviations from the mean: 0.1698µand0.1698µ+. Two different sketches are provided below. 223
Sample meanNormal density curvemu+0.1698mu+0.1132mu+0.0566mumu-0.0566mu-0.1132mu-0.16980.200.150.100.050.00(c) 2 standard deviations; . (d) About 95% (by the 68-95-99.7 rule). (e) The confidence intervals drawn will vary, but they should all have length. See the sketch above on the right. 0.1132m±20.2264m±10.5 (a) 51%or (48%, 54%). (b) 51% is a statistic from one sample. A different sample may give us a totally different answer. When taking samples from a population, not every sample of adults will give the same result, leading to sampling variability. A margin of error allows for this variability. (c) “95% confidence” means that this interval was found using a procedure that produces correct results (i.e., includes the true population proportion) 95% of the time. (d) Survey errors such as undercoverage or non-response, depending on how the Gallup Poll was taken, could affect the accuracy. For example, voluntary response will not base the sample on the population. 3%±10.6 (a) Let Gxdenote the sample mean for the 10 girls and Bxdenote the sample mean for the 7 boys. The distribution of GxxBis Normal with mean 54.554.10.4GBxxµ==and standard deviation 222.72.41.25107GBxxσ=+±. Thus, ()()00.40(0.32)0.62551.25GBGBP xxP xxPzP z>=>=>=> −=

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