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.  

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!