The candy company that makes m& m's claims that 10% of the m& m's it produces is green. suppose that the candies are packaged at random in large bags of 200 m& m's. when we randomly pick a bag of m& m's, we may assume that this represents a simple random sample of size n = 200. suppose we wish to test h0: p = 0.10 versus ha: p ≠ 0.10. suppose that in the randomly selected bag of m& m's there are only 12 green m& m's. calculate the value of the large-sample z statistic. if a 90% confidence interval were also calculated from the data, would it contain the value 0.10?

a)  

b) The 90% confidence interval would be given by (0.0324;0.0875)

So the confidence interval not contains the 0.1.  

Step-by-step explanation:

1) Data given and notation n  

n=200 represent the random sample taken

X=12 represent the number of M&M's green

estimated proportion of M&M's green.

is the value that we want to test

represent the significance level  

z would represent the statistic (variable of interest)

represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of  M&M's green is 0.1. The system of hypothesis are:  

Null hypothesis:  

Alternative hypothesis:  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value .

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

4) Statistical decision  

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level is not provided, but we can assume . The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

So based on the p value obtained and using the significance level assumed we have so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the true proportion of M&M's green is not significantly different from 0.1 or 10% .  

5) Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by and . And the critical values would be given by:

The confidence interval for the mean is given by the following formula:  

If we replace the values obtained we got:

The 90% confidence interval would be given by (0.0324;0.0875)

So the confidence interval not contains the 0.1.  

answer: 193

explanation:

200-7 = 193

believe it or not anna is correct good job anna give me a fist bump there u go gal