NEED PART F AND G ONLY https://www.chegg.com/homework-help/questions-and-answers

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NEED PART F AND G ONLY

https://www.chegg.com/homework-help/questions-and-answers/past-experience-indicates-standard-deviation-yield-manufacturing-process-3--past-6-days-pl-q26890015

Past experience indicates that the standard deviation of yield in a manufacturing process is 3%. The past 6 days of plant operation have resulted in the following yields: 91.5, 90.5, 88.7, 90.8, 89.9, and 91.4%. Is there evidence that the mean yield is not 90%? Use = 0.05. a. Write the appropriate hypothesis. b. Use P-value approach to test the hypothesis c. Use z-test to test the hypothesis d. Use confidence interval to test the hypothesis. e. What is your conclusion? If the true mean yield is 92%, what is the type II error? ize needed to recogni probability

Explanation / Answer

f. Standard deviation of yield = 3%

Here True mean yield = 92%

Here type II error is when null hypothesis is false but we failed to reject here so in this case the true mean yeild is 92% but we failed to reject the null hypothesis that mean yield is 90%.

g. If we assume that minimum sample size is n.

Here we will reject the null hypothesis, if the sample mean will be greater than the critical value

x > H + Z95% (/n)

x > 90% + 1.96 * 3/n

x > 90% + 5.88%/n

standard error of sample mean = 3%/n

Pr(x > 90% + 5.88%/n ; 92% ; 3%/n ) = 0.95

1 - Pr(x < 90% + 5.88%/n ; 92% ; 3%/n ) = 0.95

Pr(x < 90% + 5.88%/n ; 92% ; 3%/n ) = 0.05

Z = -1.645

( 90% + 5.88%/n -  92%)/ (3%/n) = -1.645

5.88%/n - 2% = -1.645 * 3%/n

10.815%/n = 2%

n = 5.4075

n = 29.24

n = 30

so minimum sample size is 30.

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