Hypothesis testing

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almill21

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A coin operated coffee machine is set to pour 8oz per cup. A random sample of the weights of a number of cups is as follows: 8.4, 8.25, 8.05, 7.84, 7.36, 8.54, 7.56, 7.56, 8.02, 7.39, 8.34, 8.26. Test the hypothesis that the machine is delivering at the level set by the manufacturer. Use a 0.05 level of significance.

Answer (need to confirm if correct or what im doing wrong)

null : mean=8
alternative: mean not equal 8

mean=95.57/12=7.964

therefore fail to reject null
 
almill21 said:
A coin operated coffee machine is set to pour 8oz per cup. A random sample of the weights of a number of cups is as follows: 8.4, 8.25, 8.05, 7.84, 7.36, 8.54, 7.56, 7.56, 8.02, 7.39, 8.34, 8.26. Test the hypothesis that the machine is delivering at the level set by the manufacturer. Use a 0.05 level of significance.

Answer (need to confirm if correct or what im doing wrong)

null : mean=8
alternative: mean not equal 8

mean=95.57/12=7.964

therefore fail to reject null
Click to expand...
Yes, that is the mean but you have not shown that it is within a "0.05 level of significance".
 
almill21 said:
Can you show how to do this exactly? I am not sure. Thank you.
Click to expand...
The next step is to find the standard deviation of the experimental data.

Use that value as an estimator of the standard deviation of the population - that is, the null hypothesis is that the mean is 8 and the standard deviation is ..

Since you have a sample of size 12, what is the standard deviation of the distribution of sample means? Use a normal distribution with that standard deviation to find the confidence interval.
 
DrPhil said:
The next step is to find the standard deviation of the experimental data.

Use that value as an estimator of the standard deviation of the population - that is, the null hypothesis is that the mean is 8 and the standard deviation is ..

Since you have a sample of size 12, what is the standard deviation of the distribution of sample means? Use a normal distribution with that standard deviation to find the confidence interval.
Click to expand...
i believe i have to compute a z score instead of t score. i dont know how to calculate the standard deviation/variance for this though. i think i know but its a rahter long process is there something im missing?
 
almill21 said:
i believe i have to compute a z score instead of t score. i dont know how to calculate the standard deviation/variance for this though. i think i know but its a rahter long process is there something im missing?
Click to expand...

Some of the calculators can directly calculate std. dev. - without you doing the calculation.

Other option is to use a software like ms-excel to calculate these statistics.
 
almill21 said:
i believe i have to compute a z score instead of t score. i dont know how to calculate the standard deviation/variance for this though. i think i know but its a rahter long process is there something im missing?
Click to expand...
The Variance is the mean of the squares minus the square of the mean:

\(\displaystyle \displaystyle V[x] = \sigma_x^2 = \dfrac 1n \sum_{i=1}^n\left( x_i^2 \right) - \left[ \dfrac 1n \sum_{i=1}^n\left( x_i \right) \right]^2\)

If you already know the mean, \(\displaystyle \mu\), then another way to calculate is as the mean of the squares of the deviations from the mean:

\(\displaystyle \displaystyle V[x] = \sigma_x^2 = \dfrac 1n \sum_{i=1}^n\left( (x_i - \mu)^2 \right)\)

Since your null hypothesis is \(\displaystyle \mu = 8\), you can more easily use the second form. Just add up the squares of the differences of each datum from 8, and divide that sum by 12. (And of course take \(\displaystyle \sqrt{V} = \sigma\).)

The observed standard deviation of the data is an estimator of the "true" standard deviation of the population. By the Sampling Theorem, the standard deviation of the sample means is \(\displaystyle \sigma / \sqrt{N}\). That is the distribution from which you need the z-score.
 
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