Need help with second set of my homework 1 When constructing

Need help with second set of my homework:

1. When constructing a confidence interval for a population mean from a sample with size 17, the number of degrees of freedom for the critical value is __? 16? 1? 17? 18?

2. A company has developed a new type of light bulb, and wants to estimate its mean lifetime. A simple random sample of 12 bulbs has a sample mean lifetime of 739 hours with a sample standard deviation of 67 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 the nearest integer

3. 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 ws 198 with a sample standard deviation of 222. 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.

4. A random sample of 13 DVD movies had a mean length of 106.4 minutes, with a standard deviation of 66.4 minutes. Find the lower bound of the 90% confidence inteval for the true mean length ofall Hollywood movies. Assume movie lengths to be approximately normally distributed. Round one decimal place.

5. 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, 19.7, 20.5, 21.8. Find the lower bound of the 95% confidence interval for the true mineral content. Round three decimal places.

6. Following are intrest 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, 4.585, 4.365, 4.774, 4.842. Find the upper bound of the 99% confidence interval for the mean rate. Round three decimal places.

Solution

1.

df = n - 1 = 17 - 1 = 16 [ANSWER, 16]

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2.

Note that              
          
Lower Bound = X - t(alpha) * s / sqrt(n)              
      
              
where              
alpha = (1 - confidence level) =    0.05          
X = sample mean =    739          
t(alpha) = critical t for the confidence interval =    1.795884819          
s = sample standard deviation =    67          
n = sample size =    12          

df = n - 1 =    11          

Thus,              

Lower bound =    704.2653715 = 704   [ANSWER]      

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