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?

Answer:a) [tex]z=\frac{0.06-0.1}{\sqrt{\frac{0.1(1-0.1)}{200}}}=-1.886[/tex]  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 takenX=12 represent the number of M&M's green[tex]\hat p=\frac{12}{200}=0.06[/tex] estimated proportion of M&M's green.[tex]p_o=0.1[/tex] is the value that we want to test[tex]\alpha[/tex] represent the significance level  z would represent the statistic (variable of interest)[tex]p_v[/tex] 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:[tex]p=0.1[/tex]  Alternative hypothesis:[tex]p \neq 0.1[/tex]  When we conduct a proportion test we need to use the z statistic, and the is given by:  [tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].3) Calculate the statistic  Since we have all the info requires we can replace in formula (1) like this:  [tex]z=\frac{0.06-0.1}{\sqrt{\frac{0.1(1-0.1)}{200}}}=-1.886[/tex]  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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  Since is a bilateral test the p value would be:  [tex]p_v =2*P(z<-1.886)=0.059[/tex]  So based on the p value obtained and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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 intervalIn 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 [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical values would be given by:[tex]t_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64[/tex]The confidence interval for the mean is given by the following formula:  [tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]If we replace the values obtained we got:[tex]0.06 - 1.64\sqrt{\frac{0.06(1-0.06)}{200}}=0.0324[/tex][tex]0.06 + 1.64\sqrt{\frac{0.06(1-0.06)}{200}}=0.0875[/tex]The 90% confidence interval would be given by (0.0324;0.0875)So the confidence interval not contains the 0.1.