# 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.

embizze | High School Teacher | (Level 1) Educator Emeritus

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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`

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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

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(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.

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

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(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`

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