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?

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 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 [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.