Thread: Confidence interval problems?

Hi all,

I really hate Statistics since the problems are very difficult





For the 9th problem:

The sample mean is (20+27+15+20+16+22+13)/7=19

The sample variance = E[X²] - 19²

= (400 + 729 + 225 + 400 + 256 + 484 + 169) / 7 - 361

= 19.428.

Bessel's correction >> 19,428 . 7/6 = 22.666

standard deviation = 4.76

... but how to go on?




For 8th problem:

Distribution binomial

mean = np = 13

standard deviation = [np (1-p)] (square root) = 3.486

... but how to go on?



I could not solve the 7th one.

You are taking statistics- don't you know what a normal distribution [b]is[b]. First consult a table of the standard normal distribution to see how many standard deviations from the mean lie within 95% or 99%. Then use the mean and standard deviation you have calculated to find the numbers you want. The table at Standard Normal Distribution Table tells me that 95% of the standard normal distribution is within 1.91 standard deviations of the mean. Since the standard deviation, in the last problem, is 4.76, 95% will be within (1.91)(4.76), or approximately 9, of the mean. Since the mean is 19, that is from 19- 9= 10 to 19+ 9= 28.

Originally Posted by HallsofIvy

You are taking statistics- don't you know what a normal distribution [b]is[b]. First consult a table of the standard normal distribution to see how many standard deviations from the mean lie within 95% or 99%. Then use the mean and standard deviation you have calculated to find the numbers you want. The table at Standard Normal Distribution Table tells me that 95% of the standard normal distribution is within 1.91 standard deviations of the mean. Since the standard deviation, in the last problem, is 4.76, 95% will be within (1.91)(4.76), or approximately 9, of the mean. Since the mean is 19, that is from 19- 9= 10 to 19+ 9= 28.

Problems 7 and 9 are small sample questions so the critical values from the t-distribution should be used with $\nu=n-1$, also the sd should be estimated with the $\frac{1}{n-1}$ formula not the $\frac{1}{n}$ formula (which is what Bessel's correction does).

.

7) For a sampling distribution the mean is the same and the standard deviation becomes 64/sqrt(15)
a 99% CI for the mean would be 912+/-2.9768*64/sqrt(15)=
912+/-49.1908=(862.81, 961.19) (2.9768 is t_alpha/2 with 14 degrees of freedom)

8)13/200=.065
a 95% CI for the proportion of defective light bulbs would be
.065+/-1.96*sqrt(.065(1-.065)/200)=
.065+/-.034167=
(.0308, .0992)

9) Assuming your standard deviation and mean are correct, 2.4469 is the t value corresponding to alpha/2 with 6 degrees of freedom
since the sample size is small and the population standard deviation is unknown, we must use the t distribution (even though the sample came from a normal distribution).
19+/-2.4469*4.76/sqrt(7)=
19+/-4.4022=
(14.60, 23.40)