Question 1139608: A machine in the lodge at a ski resort dispenses a hot chocolate drink. The average cup of hot chocolate is supposed to contain 7.75 ounces. We may assume that x has a normal distribution with σ = 0.3 ounces. A random sample of 16 cups of hot chocolate from this machine had a mean content of 7.62 ounces. Do you think the machine needs an adjustment? Use a 5% level of significance and test whether the mean amount of liquid is less than 7.75 ounces.
I have no idea how to find a p-value for problems like this.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Null Hypothesis: mu = 7.75
Alternate Hypothesis: mu < 7.75
This is a left tailed test.
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Because we know the population standard deviation (sigma), we can use the standard normal Z distribution. First we need to convert from a raw score to a z score.
The raw score is xbar = 7.62 which is the mean of the sample of n = 16 cups. We'll also plug in mu = 7.75 and sigma = 0.3
So the z score we get is...
z = (xbar - mu)/(sigma/sqrt(n))
z = (7.62-7.75)/(0.3/sqrt(16))
z = (7.62-7.75)/(0.3/4)
z = (7.62-7.75)/0.075
z = -0.13/0.075
z = -1.733333333
z = -1.73
From here, we need to find the area under the standard normal bell curve such that this area is also to the left of z = -1.13; in a visual summary, we want the shaded area of the red region below. This area will be equal to the p-value.
Note: this area is some value between 0 and 1 (excluding 0; excluding 1)
To find this area by hand would be nearly impossible or would take a long long time. So instead we rely on tables or a calculator. I'll go over both methods
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Method #1: Table
In the back of your textbook should be a table of Z values that help determine the area to the left (or right depending on your book) of the z score. I'm using this table which has areas to the left of a z score.
If you want to use the table I linked, then you'll go to the row that starts with "-1.7" and highlight everything in this row. I have done so using light blue highlighter. See the diagram below. At the same time, I've also marked the column that has 0.03 at the top with a red box. Combining the "-1.7" row and "0.03" column forms the lookup entry for z = -1.73.
The value in this row and column combination is 0.04182
This means that the area under the bell curve to the left of z = -1.73 is approximately 0.04182
Because we're dealing with a left tailed test, we don't need to do any more work beyond this point. We have our p-value and it is approximately 0.04182
Note: the p-value = 0.04182 is less than the significance level alpha = 0.05, so we reject the null. Think of the p-value as the chances of the null being correct. This line of thinking isn't entirely 100% accurate but it helps remember what to do in terms of rejecting or failing to reject the null.
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Method #2: Calculator
This method is preferred because calculators are so widespread these days that it seems silly to use a table sometimes. Though I can see the reasoning behind it. With a calculator however, we get faster more accurate results. I'll assume you are using a TI83 or TI84 or similar calculator.
On your TI calculator follow these steps
- Clear the home screen
- Hit the key labeled 2ND and then hit the VARS key
- Scroll down to normalcdf( which is the second item and hit the Enter key
- Type in -99,-1.73) then hit Enter
(the -99 can be any large negative number you want. It doesn't matter really as the area under the curve is so tiny at this point that it's practically like we're heading off to negative infinity)
You should have this result
Which is a more accurate version of the previous answer we got.
If you don't have a TI calculator, there are plenty of free online ones to use. Many of which don't require any downloads, installation, signing up, logging in, etc etc. One such calculator is this one. This is a very handy tool to not only see the numeric result, but also the visual representation of the area under the curve. The reason why I listed the TI version first is because this is the most commonly used calculator that I've come across in terms of high school and college math stats classes. Your area might be different of course.
To use the second calculator I linked, you simply click the "below" radio button, type in the z score -1.73 next to this radio button, and then click on "recalculate".
Unfortunately we don't have the luxury of getting as many decimal places as the TI calculator reports, but it's still a fairly good approximation of the p-value.
note: because we're in the standard normal Z distribution, you dont have to change the mean or standard deviation for this calculator. You can use them if you didn't convert to a z score, but that's for another lesson. So leave these values at 0 and 1 respectively.
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In summary, the p-value is approximately 0.0418 which is less than the alpha = 0.05 significance level. This indicates we reject the null hypothesis. The interpretation would be that the mean amount of hot chocolate dispensed is less than 7.75 ounces. It appears that this machine may need adjustment.
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