The masses, in grams, of apples from a certain farm have mean $$\(\mu\)$$ and standard deviation 5.2. The farmer says that the value of $$\(\mu\)$$ is 64.6. A quality control inspector claims that the value of $$\(\mu\)$$ is actually less than 64.6. In order to test his claim he chooses a random sample of 100 apples from the farm. Later another test of the same hypotheses at the $$\(2.5 \%\)$$ significance level, with another random sample of 100 apples from the same farm, is carried out. Given that the value of $$\(\mu\)$$ is in fact $$\(62.7\)$$, calculate the probability of a Type II error. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No:9709_w21_qp_63 Year:2021 Question No:7(b)
Answer:
\(\frac{m-64.6}{5.2 \div \sqrt{100}}=-1.96\)
\( m=63.5808 \)
\(\frac{63.5808-62.7}{5.2 \div \sqrt{100}}[=1.694]\)
\(1-\Phi(' 1.694\) ' \()\)
\(0.0451\)
\( m=63.5808 \)
\(\frac{63.5808-62.7}{5.2 \div \sqrt{100}}[=1.694]\)
\(1-\Phi(' 1.694\) ' \()\)
\(0.0451\)
Knowledge points:
6.5.5 calculate the probabilities of making Type I and Type II errors in specific situations involving tests based on a normal distribution or direct evaluation of binomial or Poisson probabilities.
Solution:
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