Please help confused Question 4 1 point) A company has developed a new type of l

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Question 4 1 point) A company has developed a new type of light bulb, and wants to estimate its mean lifetime. A simple random sample of 12 bulbs had a sample mean lifetime of 812 hours with a sample standard deviation of 56 hours. It is reasonable to believe that the population is approximately normal. Find the lower bound of the 95% confidence interval for the population mean lifetime of all bulbs manufactured by this new p rocess Round to the nearest integer. Write only a number as your answer. Do not write any units Your Answer Question 5 (1 polnt A random sample of 40 videos posted to YouTube was selected A month later the number of times that each had been viewed was tabulated. The mean number of viewings was 187 with a sample standard deviation of 203. Find the upper bound of the 99% confidence interval for the mean number of times videos posted to YouTube have been viewed in the first month Round to the nearest integer. Write only a number as your answer. Do not write any units Your Answer Question 6 (1 polnt A random sample of 13 DVD movies had a mean length of 125.6 minutes, with a standard deviation of 65.6 minutes. Find the lower bound of the 90% confidence interval for the true mean length of all Hollywood movies. Assume movie lengths to be approximately normally distributed Round to one decimal place (for example: 108.1). Write only a number as your answer. Do not write any units Your Answer

Explanation / Answer

4)

CI for = 95%
n = 12
mean = 812
t-value of 95% CI = 2.2010
std. dev. = 56
SE = std.dev./sqrt(n) = 16.16581
ME = t*SE = 35.58070
Lower Bound = Mean - ME = 776.41930

95% CI = 777

5)
CI for = 99%
n = 40
mean = 187
z-value of 99% CI = 2.5758
std. dev. = 203
SE = std.dev./sqrt(n) = 32.09712
ME = z*SE = 82.67670

Upper Bound = Mean + ME = 269.6767
99% CI = 270

6)

CI for = 90%
n = 13
mean = 125.6
t-value of 90% CI = 1.7823
std. dev. = 65.6
SE = std.dev./sqrt(n) = 18.19417
ME = t*SE = 32.42724
Lower Bound = Mean - ME = 93.17276

90% CI = 93.2

7)

CI for = 95%
n = 6
mean = 20.167
t-value of 95% CI = 2.5706
std. dev. = 0.979
SE = std.dev./sqrt(n) = 0.39968
ME = t*SE = 1.02740
Lower Bound = Mean - ME = 19.13960

95% CI = 19.139

8)

CI for = 99%
n = 7
mean = 4.556
t-value of 99% CI = 3.7074
std. dev. = 0.1886
SE = std.dev./sqrt(n) = 0.07128
ME = t*SE = 0.26428
Upper Bound = Mean + ME = 4.82028

99% CI = 4.820

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