HW 6

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STAT 1000

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Statistics

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Feb 20, 2024

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You must use software to solve parts of these problems. Be sure to read the questions carefully and answer all parts. In order for responses to be graded there must be both output and typed interpretation of the output in the answer. Ch 6 homework problems (5 points each) 6.14 Total sleep time of college students. In Example 5.4 (page 282), the total sleep time per night among college students was approximately Normally distributed with mean μ=7.13 hours and standard deviation σ=1.67 hours. Consider an SRS of size n=175 from this population. What is the standard deviation of the sample mean in hours? In minutes? SD of mean= 0.1262401 hours SD of mean= 7.574406 minutes Use the 95 part of the 68–95–99.7 rule to describe the variability of this sample mean. Mean= 7.13hours 95% data range= u+ 2SD / u-2SD 95% of data entries will SD= 0.126 hours =6.878, 7.382 fall within 6.878 and 7.382 What is the probability that your sample average will be below 7.0 hours? p= 0.1515558 6.42 Who is the author. 6.42 Who is the author? Statistics can help decide the authorship of literary works. Sonnets by a certain Elizabethan poet are known to contain an average of μ=8.9 new words (words not used in the poet’s other works). The standard deviation of the number of new words is σ=2.5. Now a manuscript with six new sonnets has come to light, and scholars are debating whether it is the poet’s work. The new sonnets contain an average of x¯=10.2 words not used in the poet’s known works. We expect poems by another author to contain more new words, so to see if we have evidence that the new sonnets are not by our poet, we test H0: μ=8.9 Ha: μ>8.9 Give the z test statistic and its P-value. What do you conclude about the authorship of the new poems? Z= 1.273735 P= 0.1013787 Since p is greater than 0.05 we fail to reject Ho. We fail to reject that the new sonnets are not by our poet.

6.98 Effect of sample size on significance. Complete all parts and do the final plots of 1) test statistic vs sample size and 2) p-value vs sample size. 6.98 Effect of sample size on significance. You are testing the null hypothesis that μ=0 versus the alternative μ>0 using α=0.05. Assume that σ=16. Suppose that x¯=8 and n=10. Calculate the test statistic and its P-value. Repeat assuming the same value of x¯ but with n=20. Do the same for sample sizes of 30, 40, and 50. Plot the values of the test statistic versus the sample size. Do the same for the P-values. Summarize what this demonstration shows about the effect of the sample size on significance testing. vn vz vp 1 10 1.58 0.0571 p>.05 fail to reject Ho 2 20 2.24 0.0125 p<.05 reject Ho 3 30 2.74 0.0031 p<.05 reject Ho 4 40 3.16 0.0008 p<.05 reject Ho 5 50 3.54 0.0002 p<.05 reject Ho n and z have a positive linear relationship, as z increases n increases by the same factors. This can be understood and explained by the equation z= (xbar- u) / (sigma/ sqrt n) as n changes of a factor it directly changes z. p and n have a negative exponential relationship, as n increases p decreases at first by a large factor and then the decreases become more steady/ linear as n reaches a certain point in this case observably n=30. This can be explained by the central limit theorem, as the sample size increases the values become closer to the measures of the population. In this case we can see that this distribution has a very low probability of occurring and would’ve been marked as fail to reject if the sample size was kept at 10. As we increase the size, the probability of occurring continues to decrease so we can see with a

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