1164 Words5 Pages

Chapter 7, question 13:
I would rather have a random sample of 1000 people because it is more likely to follow a normal population curve and statistically, it would have a larger number of people who are taller than 65 inches.
As stated in the question, people in a large population average 60 inches tall. So, in a normal population, the height of people drawn would be a bell curve with 60 inches being in the middle of the curve. Then, we would get people who are taller or shorter than 60 inches, as you move away from the middle. The rules of the bell curve are that 68% of the population would be within 1 standard deviation of 60 inches, 95% would be within 2 standard deviations, and 99% would be within 3 standard deviations.
Now, with 100*…show more content…*

Estimate average with 95% confidence: 1.96*30000/sqrt(n)= 4000 (1.96, for α/2=0.025 n= 217 α=0.01. Estimate average with 99% confidence 2.58*30000/sqrt(n)= 4000 1.96, for α/2=0.025 n= 375 Chapter 9, question 1: The department head does not have a good defense because the calculated probability is so low that it's more likely that this particular department has an absentee rate higher than the corporation as a whole. z = (x - mean) / (st. dev/square root of sample size) z = 4.006: x = 12 days absent mean = 8.2 days absent st. dev = 6 days absent sample size = 40 people z = (x - mean) / (st. dev/square root of sample size) z = (12 - 8.2) / (6/sqrt(40)) z = 3.8 / (6/6.32) z = 3.8 / 0.949 z =*…show more content…*

Assuming that all of the departments have the same mean and standard deviation as the corporation as a whole, there is a probability of only 0.003% that a sample of 40 people would have a mean of 12 days or more. Chapter 9, question 3: If people did not prefer one brand over another, you would predict that the number of people who answer the branded drink would be 50%. Instead, it is 70%. Because the p-value of p = 0.02 is so low, the claim that people do not prefer one brand over the other must be rejected. We can assume that customers prefer the well-known brand over the generic brand. Chapter 9, question 18: In order to have at least 8000 valid signatures, the proportion of invalid signatures must be at most .20. The significance level of the test is α =.5 / 16/100 are invalid. The estimate of is the sample proportion p P = x/n = 0.16 x: 16 n = 100 It’s the probability that a randomly selected observation from a standard normal distribution would be “as extreme as or more extreme than” the value of the test statistic. P-value = -1.000) = 0.1587 / Reject if: p-value <

Estimate average with 95% confidence: 1.96*30000/sqrt(n)= 4000 (1.96, for α/2=0.025 n= 217 α=0.01. Estimate average with 99% confidence 2.58*30000/sqrt(n)= 4000 1.96, for α/2=0.025 n= 375 Chapter 9, question 1: The department head does not have a good defense because the calculated probability is so low that it's more likely that this particular department has an absentee rate higher than the corporation as a whole. z = (x - mean) / (st. dev/square root of sample size) z = 4.006: x = 12 days absent mean = 8.2 days absent st. dev = 6 days absent sample size = 40 people z = (x - mean) / (st. dev/square root of sample size) z = (12 - 8.2) / (6/sqrt(40)) z = 3.8 / (6/6.32) z = 3.8 / 0.949 z =

Assuming that all of the departments have the same mean and standard deviation as the corporation as a whole, there is a probability of only 0.003% that a sample of 40 people would have a mean of 12 days or more. Chapter 9, question 3: If people did not prefer one brand over another, you would predict that the number of people who answer the branded drink would be 50%. Instead, it is 70%. Because the p-value of p = 0.02 is so low, the claim that people do not prefer one brand over the other must be rejected. We can assume that customers prefer the well-known brand over the generic brand. Chapter 9, question 18: In order to have at least 8000 valid signatures, the proportion of invalid signatures must be at most .20. The significance level of the test is α =.5 / 16/100 are invalid. The estimate of is the sample proportion p P = x/n = 0.16 x: 16 n = 100 It’s the probability that a randomly selected observation from a standard normal distribution would be “as extreme as or more extreme than” the value of the test statistic. P-value = -1.000) = 0.1587 / Reject if: p-value <

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