Chapter 7 Section 1 Homework Set A


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1 Chapter 7 Section 1 Homework Set A 7.15 Finding the critical value t *. What critical value t * from Table D (use software, go to the web and type t distribution applet) should be used to calculate the margin of error for a confidence interval for the mean of the population in each of the following situations? (a) A 95% confidence interval based on n = 15 observations. Using Excel I type =tinv(0.05, 14) and I get, 2.145, try it! Using applet at I enter the value which one of the tails, the left (area left of) and I hit compute! And bingo, you get the picture on the right. (b) A 95% confidence interval from an SRS of 25 observations. (c) A 90% confidence interval from a sample of size 25. On Excel I type =tinv(0.1, 24) and I get t * = Using the applet I mentioned in part (a) (d) These cases illustrate how the size of the margin of error depends upon the confidence level and the sample size. Summarize these relationships. As the sample size increases the value of t * decreases getting closer to the value of Z *
2 7.16 Distribution of the t statistic. Assume a sample size of n = 20. Draw a picture of the distribution of the t statistic under the null hypothesis. Use Table D (use software, like CrunchIt!! as well) and your picture to illustrate the values of the test statistic that would lead to rejection of the null hypothesis at the 5% level for a two sided alternative. I did this using applet at applet at program. setting the degrees of freedom to 19, and the Paint 7.19 More on one sided versus two sided P values. Suppose that x = 15.3 in the selling of the previous exercise. Would this change your P value? Use a sketch of the distribution of the test statistic under the null hypothesis to illustrate and explain your answer. This is problem Computer software reports x = 15.3 and P = 0.04 for a test of Ho: µ = 0 versus Ha: µ 0. Based on prior knowledge, you can justify testing the alternative Ha: µ > 0. What is the P value for your significance test? P( x < 15.3) = 0.02, thus the pvalue is 2(0.02) = If x = 15.3 then all that would change would be the direction of the effect. The p value in a two sided scenario would be the same; If we went to one sided where Ha: µ > 0, then the p value would be calculated by P( x > 15.3) = 0.98.
3 7.20 A one sample t test. The one sample t statistic for testing H 0 : µ = 10 H a : µ >10 from a sample of n =20 observations has the value t = (a) What are the degrees of freedom for this statistic? Degrees of freedom is 19. (b) Give the two critical values t* from Table D that bracket t. At 19 d.f < 2.10 < (c) Between what two values does the P value of the test fall? At 19 d.f < P(T > 2.10) < (d) Is the value t =2.10 significant at the 5% level? Is it significant at the 1% level? At the 5% significance t > 2.10 is statistically significant, since p value < < 0.05, but it is not statistically significant at 1%, since the p value > (e) If you have software available, find the exact P value Another onesample t test. The onesample t statistic for testing H 0 : µ = 60 H a : µ 60 from a sample of n = 24 observations has the value t = (a) What are the degrees of freedom for t? (b) Locate the two critical values t* from Table D that bracket t. (c) Between what two values does the Pvalue of the test fall (From Table D)? (d) Is the value t = 2.40 statistically significant at the 5% level? At the 1% level?
4 (e) If you have software available, find the exact Pvalue 7.25 Random distribution of trees t test. A study of 584 longleaf pine trees in the Wade Tract in Thomas County, Georgia, is described in Example 6.1 (page 354). For each tree in the tract, the researchers measured the diameter at breast height (DBH). This is the diameter of the tree at 4.5 feet and the units are centimeters (cm). Only trees with DBH greater than 1.5 cm were sampled. Here are the diameters of a random sample of 40 of these trees: Data is available at CrunchIt! site: bcs.whfreeman.com/crunchit/ips6e On the left side of the worksheet page you will see a links to chapters; click on chapter 7, then select Exercise 25. (a) Use a histogram or stemplot and a boxplot to examine the distribution of DBHs. Include a Normal quantile plot if you have the necessary software. Write a careful description of the distribution. Variable: diameter 0 : : : : : : : 9 The boxplot, two histograms, and the stemplot suggest that the distribution is right skewed. I altered the second histogram to remove some of the detail and focus on the overall distribution shape. The Normal Quantile plot also confirms that the distribution is not normal.
5 (b) Is it appropriate to use the methods of this section to find a 95% confidence interval for the mean DBH of all trees in the Wade Tract? Explain why or why not (Hint: Look at normal quantile plot). The distribution of our population is not normal, but it does not have extreme outliers as is shown by the modified boxplot. Our sample size is 40, thus the Central Limit Theorem suggests that the distribution of the sample mean will be close enough to normal so that our probability calculations are accurate enough. (c) Report the mean with the margin of error and the confidence interval. Write a short summary describing the meaning of the confidence interval. Summary statistics: Column n Mean Variance Std. Dev. Std. Err. Median Range Min Max Q1 Q3 diameter
6 The degrees of freedom is 39, thus t * = ± ± (2.8000) (21.63, 32.95) Thus, we are 95% certain that our interval contains the true mean breast height diameter. (d) Do you think these results would apply to other similar trees in the same area? Give reasons for your answer Do you feel lucky? Children in a psychology study were asked to solve some puzzles and were then given feedback on their performance. Then they were asked to rate how luck played a role in determining their scores. 10 This variable was recorded on a 1 to 10 scale with 1 corresponding to very lucky and 10 corresponding to very unlucky. Here are the scores for 60 children: Data is available at CrunchIt! site: bcs.whfreeman.com/crunchit/ips6e On the left side of the worksheet page you will see a links to chapters; click on chapter 7, then select Exercise 25. (a) Use graphical methods to display the distribution. Describe any unusual characteristics. Do you think that these would lead you to hesitate before using the Normalitybased methods of this section? Ok, this is obviously not normally distributed, I do not need a normal quantile plot. This is an open and shut case. The result is very fascinating in that it seems to indicate we have two set of children here; maybe there are some issues on how each set of children sees life. Thus a statistician would try to find the correct category that would separate the children into their proper groups. But that is not the task here. The question however is can we use the confidence interval formula? For this to happen the distribution of xbar must be close to normal. Is a sample of 60 close enough for this data set? If you were to say no, I would totally understand your choice. However, one item that captures my eye is the symmetry of the distribution. Because of the symmetry it may be possible that a sample of 60 is large enough.
7 (b) Give a 95% confidence interval for the mean luck score. (c) The children in this study were volunteers whose parents agreed to have them participate in the study. To what extent do you think your results would apply to all similar children in this community?
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