Help with Confidence Intervals

[Rugrats3]

The questions regarding confidence intervals and population means are hard to understand: Please help me set them up in a formula and solve.

#3 An inspector has been given the assignment to estimate the mean net weight og pakages of beef labeled 3 pounds. He realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveal the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds.

a. What is the estimated population mean?

b. Determine a 95 percent confidence interval for the population mean?




#4 The HR Department would like to include a dental plan for the benefits package. How much does a typical employee and his or her family spend per year on dental expenses? A sample of 45 employees reveals the mean amount spent last year was $1,820, with a standard deviation of $660.

a. Construct a 95% confidence interval for the population mean.

b. The information from part (a) was given to the president. He indicated he could afford $1,700 of dental expenses per employee. Is it possible that the population mean could be $1,700? Why?

[JohnM]

The questions regarding confidence intervals and population means are hard to understand: Please help me set them up in a formula and solve.

#3 An inspector has been given the assignment to estimate the mean net weight og pakages of beef labeled 3 pounds. He realizes that the weights cannot be precisely 3 pounds. A sample of 36 packages reveal the mean weight to be 3.01 pounds, with a standard deviation of 0.03 pounds.

a. What is the estimated population mean?

b. Determine a 95 percent confidence interval for the population mean?




#4 The HR Department would like to include a dental plan for the benefits package. How much does a typical employee and his or her family spend per year on dental expenses? A sample of 45 employees reveals the mean amount spent last year was $1,820, with a standard deviation of $660.

a. Construct a 95% confidence interval for the population mean.

b. The information from part (a) was given to the president. He indicated he could afford $1,700 of dental expenses per employee. Is it possible that the population mean could be $1,700? Why?

(3)a.
the estimate of the population mean is the mean of the sample - 3.01

b.
the 95% confidence interval around the mean is
sample mean +/- (Z * (s/sqrt(n))
= 3.01 +/- (1.96 * (.03/sqrt(36))

(4)a.
do the same way as 3b

b.
if the confidence interval you compute in 4a contains 1700, then yes, because over a "large" number of samples, the 95% confidence interval will contain the true population mean 95% of the time

[Rugrats3]

Thanks for your help but are you using the z or t tables at all for answers and if so, where did you look. These tables are confusing!

[quark]

The samples sizes in #3 and #4 are relatively large (>30), you can use the z table.

Below is a nice tutorial for z table:

http://www.mis.coventry.ac.uk/~styrr...s/ntabexam.htm

[JohnM]

Thanks for your help but are you using the z or t tables at all for answers and if so, where did you look. These tables are confusing!

If you draw lots of samples, but the sizes of those samples are pretty small, say below 30, the means won't really follow a normal distribution - they follow a t-distribution. With really small sample sizes, the distribution's tails are "fatter" and the peak is kind of low.

As the sample sizes increase, the shape of the distribution of the sample means will "approach" a normal curve, and when they're "large" (n >= 30), you can use the normal distribution tables (Z).