Hypothesis Testing of Mean, Proportion&Variance:M&M Candies

See attached file.

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 number of candies per bag. For the Bonus you will be using the methods of 7.5.

You can use StatCrunch to assist with the calculations. A link for StatCrunch can be found under Tools for Success in Course Home. Here is also a link: http://statcrunch.pearsoncmg.com/statcrunch/larson_les4e/dataset/index.html. You can also find additional help on both confidence intervals and StatCrunch in the Online Math Workshop under Tab: 'MAT300 Archived Workshops'. Specifically you will be looking for Hypothesis Tests and Using Technology - Hypothesis Testing.

Be sure to state clear hypotheses, test statistic values, 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.

3 pts. Test their claim that the true proportion of blue M&Ms candies is 0.24 at the 0.05 significance level.

3 pts. Test their claim that the true proportion of orange M&Ms candies is 0.20 at the 0.05 significance level.

3 pts. Test their claim that the true proportion of green M&Ms candies is 0.16 at the 0.05 significance level.

3 pts. Test their claim that the true proportion of yellow M&Ms candies is 0.14 at the 0.05 significance level.

3 pts. Test their claim that the true proportion of red M&Ms candies is 0.13 at the 0.05 significance level.

3 pts. Test their claim that the true proportion of brown M&Ms candies is 0.13 at the 0.05 significance level.

3 pts. 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 (? unknown).

BONUS: 5 pts. 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 (? = 0.05) that the true standard deviation for number of candies per 1.69 oz bag is no more than 1.5 (? < 1.5).

HELP:

Color proportions

Revisiting the purple example from before: We had 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 â?  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 test statistic, z, is positive and greater than 1.96 OR if the test statistic, z, is negative and smaller than -1.96. (NOTE: This is the same as if the absolute value of the test statistic is greater than 1.96.)

Review: â?' sample proportion (0.209143)
p â?' assumed value in null (0.21)
q â?' 1 - p (0.79)
n â?' total number of candies (3500)

Test statistic:

Because the test statistic is between -1.96 and 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 and n.

IF using StatCrunch, you will want Stat > Proportions > One Sample > with summary. In the first window, you will enter the same information as for part 3: number of the color (number of successes) and total number of candies (number of observations). Then click Next, and in the following window, enter the claimed proportion as a decimal in the box next to 'null', select the inequality that matches the alternative hypothesis and then click Calculate. The output will include the test statistic (Z-Stat) and the p-value.

Hypothesis test results:
p : proportion of successes for population
H0 : p = 0.21
HA : p â?  0.21

Proportion

Count

Total

Sample Prop.

Std. Err.

Z-Stat

P-value

p

732

3500

0.20914286

0.006884766

-0.12449848

0.9009

Mean

When you test for the mean number of candies per bag, you will need (sample mean), s (sample standard deviation) and n (total number of bags) as before.

The test statistic is a z, because we have a large sample.

Test statistic:

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

Input: Stats

m0: assumed mean value

s: sample or known standard deviation

: 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 and n.

IF using StatCrunch, you will use Stat > Z Statistics > one sample as with part 3.

BONUS

This is a test about a standard deviation. You can use StatCrunch, however, StatCrunch deals with variances, so you would enter 1.5² = 2.25 as the null value. You would use Stat > Variance > One Sample. Again, it would be best to use the actual data entered into a column, but you can also use summary values, as long as you carry at least 4 decimal places.