The candy company that makes M&M;\'s claims that 10% random, and the small bags

Question

The candy company that makes M&M;'s claims that 10% random, and the small bags contain 55 M&M;'s. When we randomly pick a bag random sample of size n 55. Suppose that in a randomly of the M&M;'s it produces are green Suppose that the candies are packaged a of M&Ms;, we may assume that this represents a simple selected small bag of M&M;'s, there are 11 green M&Ms.; At the 1% significance level are the data statistically significant for testing if the true proportion of green M&Ms; is not equal to 10%? Select one: O a. Yes O O b. No c. This cannot be determined from the information given Check

Explanation / Answer

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.10

Alternative hypothesis: P 0.10

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).

= sqrt[ P * ( 1 - P ) / n ]

= 0.041

z = (p - P) /

z = 2.44

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 2.44 or greater than 2.44.

Thus, the P-value = 0.0146

Interpret results. Since the P-value (0.0146) is greater than the significance level (0.01), we have to accept the null hypothesis.

The given data is not statistically significant for testing that the true proportion is not equal to 10%.

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