Evening everybody...I'm in the midst of studying for an upcoming final, and I'm running into some problems that weren't discussed in class in any great detail (we basically went through 2 chapters of material in less than 3 meetings and crammed everything in) due to the semester time constraints. First example question:
1. An advertisement claims that the average life of Tyre brand Tires is 50,000 miles, and the known standard deviation of this population is known to be 2,640 miles. Test this claim at the level. (Note - the mean of a random sample of 81 of these tires was 50,488 miles.
Information - So far I understand the two hypotheses to be and the . I believe since the sample has 81 tires that the distribution is normal, and I'm assuming that because of the note that the mean would be 50,488 and that the standard deviation would be the original standard deviation of 2,640 divided by the square of the number of sample trials which is the square root of 81, or 9. This would lead me to believe that for the sample distribution, but the population distribution would still be . At this point assuming I'm correct since sigma is known that I would run a ZTest, and using the calculator I end up with a P value of 1, and since the claim cannot be rejected. I'm not very confident in my calculations at this point, so any insight to this problem would be extremely helpful.
Unfortunately there's another one similar to this one that looks like a T-Test is required.
2. The weights of toys from a manufacturer's process is claimed to have a normal distribution with a mean of 7 ounces. Test This claim at the 0.01 level of significance that the process is making toys that are heavier than they are supposed to be if a random sample of 15 toys had a mean weight of 7.12 ounces and a standard deviation of 0.15.
Information - I believe that and . The population distribution would be and the sample distribution would be . Since the standard deviation of the population is unknown, a T-Test would be ran and the P value ends up being 0.00393, and since the claim can be rejected. Similar principal to the first number except for using a T-Test instead of Z-Test, and even though I have a little more confidence in this example I'm still not 100% sure.
Finally, the last problem is probably the easiest one but I'm not completely sure about my answers.
3. Find a 92% confidence interval for the true mean of a normal population whose standard deviation is known to be 12 ounces, if the mean of a random sample of size 36 was 18.6 pounds.
Information - Since the mean for the population isn't known, the only information I have is that using .75 since 12 ounces is 3/4ths of a pound. The sample distribution is because 0.75 over the square root of the n=36 is six, and .75/6 is equal to 1/8th or 0.125. I assume I use the sample data to run a Z-Interval on the calculator and come out with . Obviously I'm thrown off because of the missing mean for the population, but I'm hoping that it doesn't matter since I had enough information for the sample test that I could run the calculations and report my findings.
Any help at this point would be deeply appreciated. Thank you for your time.