# Math Help - I need HELP on how to set this up!

1. ## I need HELP on how to set this up!

A manufacturer of bolts wants to determine the length of the manufactured bolts. A random sample of 160 bolts shows that the sample mean length is 2.9 inches and that the sample standard deviation is 0.1 inches.

Part (a) (5 points)
Calculate a 95 percent confidence interval for the mean length of all the bolts (i.e., for the population mean).

Part (b) (2 points)
Consider the following statement: “It is always better to construct a 99-percent confidence interval instead of a 95-percent confidence interval.” Do you agree or disagree with this statement? Justify your answer

2. Originally Posted by nac123
A manufacturer of bolts wants to determine the length of the manufactured bolts. A random sample of 160 bolts shows that the sample mean length is 2.9 inches and that the sample standard deviation is 0.1 inches.

Part (a) (5 points)
Calculate a 95 percent confidence interval for the mean length of all the bolts (i.e., for the population mean).

Part (b) (2 points)
Consider the following statement: “It is always better to construct a 99-percent confidence interval instead of a 95-percent confidence interval.” Do you agree or disagree with this statement? Justify your answer
Part A:
For any confidence interval, call value of the confidence level $C$ (such as 90% or 95%, as in your case). Then the probability that the true population mean (called $\mu$) lies within a given interval with a confidence level of $C$ is:

$P(\bar{x} - z \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{x} + z \frac{\sigma}{\sqrt{n}})$ where $\bar{x}, z, \sigma, n$ are your mean, z-score, standard deviation, and sample size, respectively.

So your confidence interval is between the points
$\bar{x} - z \frac{\sigma}{\sqrt{n}}, \bar{x} + z \frac{\sigma}{\sqrt{n}}$

To find a given confidence interval, plug in those values. In your case, [tex]\bar{x} = 2.9, n = 160, \sigma = 0.1[tex]. The z-score will be based on your confidence level. Look them up in a z-table. A 90% confidence interval will exclude 5% of the normal distribution on neither side (a total of 10% excluded). A 95% confidence interval will exclude 2.5% (for a total of 5% excluded).

Part B:
Consider what a confidence interval means. A 90% confidence interval means that repeated trials using those values of $z, \sigma, n$ will give you intervals that contain the true mean 90% of the time. A 95% confidence will do the same thing 95% of the time. But for a 95% confidence interval, the z-value is higher, so what happens to the width of the interval?

The answer is that both have their pros and cons - you are essentially trading off how certain you are about where the mean is with precise of a net you are casting. The higher the confidence level, the wider the interval is. It's sort of like: would you prefer "I am 95% that this guy is in America" vs. "I am 80% sure this guy is in New York?"