# Thread: Conf Interval and Hypothesis Testing

1. ## Conf Interval and Hypothesis Testing

Question 1
You take a random sample of 16 from a normal population of mean 30 and population standard deviation of 8. Find the 90% confidence interval for the population mean.

Since pop mean is known i dont need t score, right?
also, do I have to divide the sd by the srt of n even though i am given the pop sd?

the answer I got was, 30 plus/minus 1.645(8/srt2)

Question 2
A purchaser of light bulbs wants to test the hypothesis that the average lifetime is more than 100 hours. She takes a random sample of 16 and gets a sample mean of 96 with a sample standard deviation of 8 hours.
a) Should she accept the hypothesis that they last more than 100 hours at the 95% level of confidence?
b) What about at the 99% level of confidence.\?

I dont know what to do with this problem. Its a small sample size but not sure what t score to use or what to do with it :P
i can get the t score of .05 which is 1.753 and i have a formula for t score thats u - x/(s/sqrt n)

i could use a lot of help with 2. im not asking for the answer, just a little info on how to do it.

Thank you.

2. For Q.1,

Since the question only asks about "confidence interval for the population mean" and you already know that the population mean is 30 and population standard deviation is 8,then, the C.I. is just:

Upper limit = 30 + 1.645 * 8
Lower limit = 30 - 1.645 * 8,

in my opinion, there is no need to divide the std by anything.

3. For question #2, what is your null hypothesis? The mean is 100?

If so, you would perform the test of mean=100 vs. mean>100.

Since you are not given the standard deviation, you must use a t-test. At the 95% significance level, you need to calculate your test statistic T = (sample mean - 100)/Sqrt[Sample variance/n]. If T > t(.05) with n-1 degrees of freedom, then you reject the null hypothesis.