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 H_a: 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) Yes
B) No
C) This cannot be determined from the information given.

Question

Mariah Brigham · Accepted Answer

Final Answer : B, Explanation :   The large-sample z statistic is calculated using the formula   z = p ^ − p 0 p 0 ( 1 − p 0 ) n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} z = n p 0 ​ ( 1 − p 0 ​ ) ​ ​ p ^ ​ − p 0 ​ ​  , where   p ^ \hat{p} p ^ ​   is the sample proportion,   p 0 p_0 p 0 ​   is the hypothesized population proportion, and   n n n   is the sample size. Here,   p ^ = 12 200 = 0.06 \hat{p} = \frac{12}{200} = 0.06 p ^ ​ = 200 12 ​ = 0.06  ,   p 0 = 0.10 p_0 = 0.10 p 0 ​ = 0.10  , and   n = 200 n = 200 n = 200  . Plugging these values into the formula gives   z = 0.06 − 0.10 0.10 ( 1 − 0.10 ) 200 = − 0.04 0.09 200 = − 0.04 0.0212 ≈ − 1.89 z = \frac{0.06 - 0.10}{\sqrt{\frac{0.10(1-0.10)}{200}}} = \frac{-0.04}{\sqrt{\frac{0.09}{200}}} = \frac{-0.04}{0.0212} \approx -1.89 z = 200 0.10 ( 1 − 0.10 ) ​ ​ 0.06 − 0.10 ​ = 200 0.09 ​ ​ − 0.04 ​ = 0.0212 − 0.04 ​ ≈ − 1.89  . This z-value indicates that the observed proportion of green M&M's (0.06) is significantly lower than the hypothesized proportion (0.10) at a 90% confidence level. Therefore, a 90% confidence interval calculated from this data would not contain the value 0.10, indicating that the true proportion of green M&M's is significantly different from 0.10.

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