Question

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1. 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 651 hours with a sample standard deviation of 31 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 process. Round to nearest integer.

2. 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 109 with a sample standard deviation of 376. 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.

3. A random sample of 13 DVD movies had a mean length of 116.6 minutes, with a standard deviation of 73.3 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.

4. Six measurements were made of the mineral content (in percent) of spinach, with the following results. It is reasonable to assume that the population is approximately normal. 19.1,20.1,20.8,20.0,20.5,20.3. Find the lower bound of the 95% confidence interval for the true mineral content. Round to three decimal places.

5. Following are interest rates (annual percentage rates) for a 30-year-fixed-rate mortgage from a sample of lenders in a certain city. It is reasonable to assume that the population is approximately normal. 4.327,4.461,4.547,4.574,4.365,4.304,4.842. Find the upper bound of the 99% confidence interval for the mean rate. Round to three decimal places.

Answer 1

Q 1) The lower bound of the 95% confidence interval for the population mean lifetime of all bulbs manufactured by this new process :

Here we apply t test because the population standard deviation is unknown

Sample mean

Sample standard deviation s =31

n = 12

degrees of freedom = n-1 = 12-1 = 11

The critical value of t for 11df at 5% Significance level is 2.20

The lower bound of CI is

Answer: 631