# I need help understanding confidence intervals and null, alternative hypothesis?A machine is programmed to put 737 grams of salt in a container. due to uncontrolled variation in the process, there...

I need help understanding confidence intervals and null, alternative hypothesis?

A machine is programmed to put 737 grams of salt in a container. due to uncontrolled variation in the process, there is a variation in content from container to container. To estimate the mean amount of salt per container, a sample of 50 boxes is selected and mean = 739.5 grams. From experience with the machine, its known standard deviation= 7.5 grams.

A. Find the 90% confindenced interval estimate for u and write the confidence statement.

B. Using the same information, from the previous problem,

How large a sample should be taken if the population mean is to be estimated with 95% confidence to within 8 grams?

2. Stat the null and alternative hypothesis for the following:

Charles heard that the mean height of professional basketball players is greater than 74 inches. He believes that the mean height of professional basketball players is less than 74 inches.

*print*Print*list*Cite

### 1 Answer

1) a) Given `sigma=7.5,bar(x)=739.5,n=50` Find the 90% confidence interval estimate for `mu` .

Since we know the population standard deviation we use a `z` test:

If the confidence level is .9 then `alpha=1-.9=.1` and `alpha/2=.05` . Using the standard normal table we find that `z_(alpha/2)=1.64` . (You can find .0500 in the table and realize we use the positive number, or find .9500 in the table)

The confidence interval is given by:

`bar(x)-z_(alpha/2)(sigma/sqrt(n))<=mu<=bar(x)+z_(alpha/2)(sigma/sqrt(n))`

`bar(x)` is the mean of the sample, `n` is the size of the sample, `z_(alpha/2)` is associated with the confidence level (see above) and `sigma/sqrt(n)` is the standard error -- the variance of the sample is likely to be smaller than the population variance.

Plugging in the numbers we get:

`739.5-1.64(7.5/sqrt(50))<=mu<=739.5+1.64(7.5/sqrt(50))`

`737.76<=mu<=741.24`

-------------------------------------------------------------------

**With 90% confidence we can claim that the average amount of salt per container is in the interval `737.76<=mu<=741.24` or `739.5+-1.74` grams**

------------------------------------------------------------------

(b) Find the sample size required for a 95% confidence level if the mean is to be found within 8 grams:

Here we use `n=((z_(alpha/2)*sigma)/E)^2` where `n` is the sample size we seek, `z_(alpha/2)` is computed as above, `sigma` is the population standard deviation, and `E` is the error. This formula is just the formula for the error, `E=z_(alpha/2) sigma/sqrt(n)` solved for `n` .

For a 95% confidence level, `alpha=.05,alpha/2=.025` and `z_(.025)=1.96`

Plugging in we get:

`n=((1.96*7.5)/8)^2~~3.37` . We round this to the next higher integer.

-----------------------------------------------------------------

**For a 95% confidence level that the sample mean is within 8 grams of the true mean, we need a sample of size 4.**

----------------------------------------------------------------

(2) In most treatments, the null hypothesis states that there is no difference between a statistic and a parameter ( or no correlation in the case of multiple variables). Thus, normally the null hypothesis is stated with an equals sign. If your book treats this differently, my apologies:

The null hypothesis is that the mean height really is 74 inches. The alternative hypothesis (the claim in this case) is that the mean height is less than 74 inches.

`H_0:mu=74` `H_1:mu<74`

**Sources:**