I am getting better at Stat - i still hate it though! I had a whole bunch of q's to do today and i did every single one of them! I was hoping that some1 could please help me check and make sure that my answers are correct! for ease of understanding I put the answers that i got in BOLD and a different color!
1. Of 139 adults selected randomly from one town, 30 of them smoke.
The main concern I have on this question is we take (30/139) = .215 or 21.5% ... now do we keep that as is or do we round it off to 22% ?
Mr F says: Use 0.216.
a. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke. 12.6%<p<30.4%
Mr F says: .
b. Construct a 95% confidence interval for the true percentage of all adults in the town that smoke. 14.7%<p<28.3%
Mr F says: .
2. The speeds (in mph) of the cars passing a certain checkpoint are measured by radar. The results for a random sample are shown below.
44.4 41.8 42.9 40.7 43.5 40.1 44.6 41.9
44.4 42.0 43.5 41.9 40.7 43.5 41.8
a. Find x-bar and s for this data set. x-bar = 42.513 s = 1.439
b. Construct a 90 percent confidence interval for the mean speed of cars passing this checkpoint. You may assume the speeds are normally distributed. 41.9<-->43.1241
Mr F says: Use the t-statistic here (15 - 1 = 14 degrees of freedom) because population standard deviation is unknown and n = 15 is small. The critical value is t = 1.345.
3.A random sample of the duration of 50 telephone calls handled by a local telephone company had a mean of 11.6 min and a standard deviation of 3.8 minutes. Find a 95% confidence interval for the true mean duration of calls handled by the company. [10-55 - 12.65]
Mr F says: Use the z-statistic here because athough population standard deviation is unknown, n = 50 is large. The critical value is z = 1.96.
4. A parcel service selected a random sample of 17 packages in order to estimate the mean weight of all packages received by the service. The sample had a mean weight of 15.3 pounds and a standard deviation of 1.9 pounds. Find the 95% confidence interval for the true mean weight of all packages received by the parcel service. You may assume package weights are normally distributed. [6.27 - 24.33] --> this range seemed to high to me! any ideas or thoughts?
Mr F says: Use the t-statistic here (17 - 1 = 16 degrees of freedom) because population standard deviation is unknown and n = 17 is small. The critical value is t = 1.746.
I cannot begin to imagine how you got your answer!
Thanks for all your help in advance and hope that these answers are right!