Problem: Goal: We want to determine the true mean score of all bowlers in the wo

Question

Problem: Goal: We want to determine the true mean score of all bowlers in the world. Below are some useful statistics that will assist you to solve the next problems. Value 192 10 16 Statistics Sample Mean () Sample Standard Deviation () Sample Size (n) Assume the Population Standard Deviation (a) is 5 and population is normally distributed 1) Compute the margin of error for a 90% confidence interval of the true t mean score of all bowlers in the world. 2) Construct a 95% confidence interval of the true t mean score of all bowlers in the world. Suppose, now, that we do not have information about the population standard deviation, but we know that the distribution score of all bowlers in the world is somewhat normally distributed Construct a 95% confidence interval for the true average score of all bowlers in the world. Hint: Use the fact that the Margin of Error is 0.2064. 3) 4) In the confidence interval above, how would you write conclusion so that everybody can understand it? A survey of 1000 houses in the U.S showed that 25% put cameras in their backyard. 5) Construct a 95% confidence interval for the true proportion of houses with cameras in their backyard

Explanation / Answer

1)

we know that margin of error is

MOE = Z*SD/sqrt(n) , for 95% Z = 1.96 from the z table

SD = 10
n = 16
putting the values in the equation

we get

MOE = 1.96*10/sqrt(16) = 4.9

2)

again in this case the CI is given as

Mean +- MOE , we just calculated the MOE of the sample as 4.9
putting the value in the equation

192+- 4.9 and solving for plus and minus sign we get

192+4.9 and 192-4.9

187.1, 196.9

3)

for the whole population we are given that the MOE = 0.2064

so we use the same equation again as
Mean +- MOE

192+0.2064 and 192-0.2064

191.79 and 192.2064

4)

we can say that we are 95% confident that the true value of the mean would lie in the interval 191.79 and 192.2064 and not outside the interval

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