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# Statistics: Testing Claims

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Use the M&Ms® data to complete this assignment. You will be using the methods of 7.4 for the color proportions and 7.2 for the mean total number of candies per bag. For the Bonus you will be using the methods of 7.5. You may submit your answers in Excel, Word or pdf format. Be sure to state clear hypotheses, test statistic, critical value or p-value, decision (reject/fail to reject), and conclusion in English. When doing calculations for the color proportions, keep at least 4-6 decimal places sample proportions, otherwise you will encounter large rounding errors.

Masterfoods USA states that their color blends were selected by conducting consumer preference tests, which indicated the assortment of colors that pleased the greatest number of people and created the most attractive overall effect. On average, they claim the following percentages of colors for M&Ms® milk chocolate candies: 24% blue, 20% orange, 16% green, 14% yellow, 13% red and 13% brown.

Test their claim that the true proportion of blue M&Ms® candies in a 1.69 oz bag is 0.24 at the 0.05 significance level.

Test their claim that the true proportion of orange M&Ms® candies in a 1.69 oz bag is 0.20 at the 0.05 significance level.

Test their claim that the true proportion of green M&Ms® candies in a 1.69 oz bag is 0.16 at the 0.05 significance level.

Test their claim that the true proportion of yellow M&Ms® candies in a 1.69 oz bag is 0.14 at the 0.05 significance level.

Test their claim that the true proportion of red M&Ms® candies in a 1.69 oz bag is 0.13 at the 0.05 significance level.

Test their claim that the true proportion of brown M&Ms® candies in a 1.69 oz bag is 0.13 at the 0.05 significance level.

On average, they claim that a 1.69 oz bag will contain more than 54 candies. Test this claim (µ > 54) at the 0.01 significance (&#963; unknown).

It is important that the total number of candies per bag does not vary very much. As a result of this quality control, the desired standard deviation is 1.5. Test the claim (&#945; = 0.05) that the true standard deviation for number of candies per 1.69 oz bag is no more than 1.5 (&#963; < 1.5).

Revisiting the purple example from before: we had found 732 purple candies out of 3500 total candies. The sample proportion of purple candies is 732/3500 = 0.2091428571.
Now let's say you want to test that the true proportion of purple candies is 21% (0.21).
First define your hypotheses: Claim - p = 0.21
H0: p = 0.21 (null)
H1: p not equal 0.21 (alternative)

Next we need to calculate the test statistic. For this type of test, it is a z and a two tailed test. You have been asked to test at alpha = 0.05, so we will reject the null if the absolute value of z is greater than 1.96 (make z positive and if it is greater than 1.96 reject).
z = (phat - p)/sqrt(p*q/n)
Review: phat -> sample proportion (0.209143)
p -> assumed value in null (0.21)
q -> 1 - p (0.79)
n -> total number of candies (3500)

Test statistic:
z = (0.209143 - 0.21)/sqrt(0.21*0.79/3500) = (-0.000857/sqrt(0.0000474)) = -0.000857/0.0068847658 = -0.1245
Absolute value of z (test statistic) = 0.1245
Because this is NOT greater than 1.96, we FAIL TO REJECT. We have insufficient evidence to suggest the true proportion is not 0.21.
You will follow this procedure for EACH color.

If using the TI 83/84: STAT, TESTS, 1-PropZTest

Po: assumed proportion (0.21)

x: number of successes (732)

n: total number of candies (3500)

In the next line, select the correct alternative hypothesis/test, then Calculate, Enter.

On the next screen, the second line shows the test.

The next line has the test statistic.

The next line has the p-value of the test (if less than significance level, reject null)

The next two lines have phat and n.

When you test for the mean total number of candies, you will need xbar, s and n as before.
The test statistic is a z, because we have a large sample.

Test statistic:
z = (sample - assumed)*sqrt(n)/s = (xbar - mu)*sqrt(n)/s

Note: This formula looks slightly different from the one given on page 381, but algebraically it is the same. It results in less mathematical errors.

IF using the TI 83/84: STAT, TESTS, Z-Test

Input: Stats

m0: assumed mean value

s: sample or known standard deviation

Xbar: sample mean

N: sample size

Then select the correct alternative hypothesis/test, then Calculate, Enter.

On the next screen, the second line shows the test.

The next line has the test statistic.

The next line has the p-value of the test (if less than significance level, reject null)

The next two lines have xbar and n.