Mathematics, 05.12.2019 17:31 peterradu47781

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?

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Answer from: 604341

a) $z=\frac{0.06-0.1}{\sqrt{\frac{0.1(1-0.1)}{200}}}=-1.886$

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

$\hat p=\frac{12}{200}=0.06$ estimated proportion of M&M's green.

p_o=0.1 is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

p_v 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: p=0.1

Alternative hypothesis: $p \neq 0.1$

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

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

3) Calculate the statistic

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

$z=\frac{0.06-0.1}{\sqrt{\frac{0.1(1-0.1)}{200}}}=-1.886$

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

p_v =2*P(z

So based on the p value obtained and using the significance level assumed $\alpha=0.05$ we have $p_v\alpha$ 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 $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical values would be given by: