Math 300 MM Project

After performing the test, a chi square value of 126.58 was determined. With a degrees of freedom value of 4, this constituted a p value of 2.101644 x 10-26. With an alpha level of .01, the null hypothesis was rejected.

2. Each bag of M&Ms has candies of several colors. Thus, the three red M&Ms that Barry took out of his bag (are/are not) a representative sample.

2. Each bag of M&Ms has candies of several colors. Thus, the three red M&Ms that Barry took out of his bag (are/are not) a representative sample.

Topics Distribution of the sample mean. Central Limit Theorem. Confidence intervals for a population mean. Confidence intervals for a population proportion. Sample size for a given confidence level and margin of error (proportions). Poll articles. Hypotheses tests for a mean, and differences in means (independent and paired samples). Sample size and power of a test. Type I and Type II errors. You will be given a table of normal probabilities. You may wish to be familiar with the follow formulae and their application.

Find the probability that the ice cream was chocolate flavor, given that it was sold in a cup. Place your answer, rounded to 4 decimal places, in the blank. For exampe, 0.3456 would be a legitimate entry. 29.4

23. The appropriate alternative hypothesis for a two-tail test to determine if mean body weight of all the men who have joined a health club is the same as 185 pounds would be

The objective of this part of the project is was to construct a 95% confidence interval for the proportions of blue, orange, green, yellow, red and brown m&ms. The results were:

Confidence intervals at 95%, the total number of candies for each color will fall between two end points, was calculated for each color. The results showed as follows: blue 19.35 to 22.09%, orange 22.29 to 24.13%, green 17.21 to 19.83%, yellow 10.57

A pharmaceutical company is testing the effectiveness of a new drug for lowering cholesterol. As part of this trial, they wish to determine whether there is a difference between the effectiveness for women and for men. Using   = .05, what is the value the test statistic?

The percentage of population which its trait is the recessive bb = (12/100,000) X 100= 0.012%.

Suppose the significance level for a hypothesis test is α = 0.05. If the p-value is 0.049, the decision is to

Compute the critical value ta/2 that corresponds to a 95% level of confidence and degree of freedom = 10.

c. Find the critical value (or values), and clearly show the rejection and non-rejection regions.

Results: The p-value of 0.00203 is p<0.05 thus we reject the null hypothesis and conclude that there is a significant difference between the two means.

Using a 10% (0.1) significant level also known as Alpha, the p-value of the two tailed test shows that it is 38% (0.38) which is obviously higher than 10% which implies that we cannot reject the null hypothesis. Ho: μ1 = μ2