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https://www.webassign.net/web/Student/Assignment-Responsesiast?dep-17355498 C Search 2. + Question Details My Notes ÷ Ask Your I would like to construct a confidence interval to locate the mean height for a population of mammals. I draw a random sample, and the 99% confidence interval for the mean height is [20,52] (for this interval the margin or error-16), I would like to construct an interval which has a margin of error-8. What can I plausibly do to obtain a margin of error-87 O Either (B) or (C) would work. O (E) Either do (A) or reduce the standard deviation of the population. O (A) Inc O (C) Since the critical values for the 99% and 80% confidence intervals are 2.56 and 1.28 respectively, we can decrease the level of confidence from a 99% level to 80% level. O (B) Increase the sample size by factor four rease the sample size by factor eight. 3. Question Details My Notes Ask Your The number of heart beats per minute of a mouse varies from mouse to mouse. Suppose that the number of beats per minute of 15 mice is measure. The sample mean for these 15 mice is 39 beats. The STANDARD ERROR for the sample mean is 0.8. Using this information (a) find the standard deviation for a the heart beat of a mouse and (b) how large a sample size do I need for the margin of error for a 95% confidence interval to be to 0.5 (use the normal distribution)? O The standard deviation is 0.2 (this is the variability in heart beats between mice). Therefore solving the equation the minimum sample size is 1. O The standard deviation is 3.1 (this is the variability in heart beats between mice). Therefore solving the equation the minimum sample size is 148 O The standard deviation is 3.1 (this is the variability in heart beats between mice). Therefore solving the equation the minimum sample size is 10 The standard deviation is 0.8 (this is the variability in heart beats between mice). Therefore solving the equation the minimum sample size is 10 The standard deviation is 12 (this is the variability in heart beats between mice). Therefore solving the equation the minimum sample size is 2213 3:17 PM 10/13/2017 2 Type here to searchExplanation / Answer
2) either B or C would work as each individually reduce margin of error by a factor of 2.
3)as std deviation =std error *(n)1/2 =0.8*(15)1/2 =3.1
also for margin of error E =0.5
and for 95% CI ; z =1.96
hence sample size =(z*std deviation/E)2 =148
hence second option is correct : std deviation is 3.1 . therefore solving the equation sample size is 148
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