..il AT&T; 7:41 PM webassign.net Assignment Submission For this assignment, you
ID: 3365636 • Letter: #
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..il AT&T; 7:41 PM webassign.net Assignment Submission For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer Assignment Scoring Your best submission for each question part is used for your score. A quality control engineer at a particular led screen manufacturer is studying the mean number of defects per screen. Based on historical evidence, the mean number of defects per screen was thought to be 2.58. There have recently been changes to the manufacturing process, and the engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58. Using the number of defects on each of 50 sample screens shown below, conduct the appropriate hypothesis test using a 0.1 level of significance. Assignment 7q1 data a) What are the appropriate null and alternative hypotheses? O Ho: -2.58 versus Hai * 2.58 O Ho: x = 2.58 versus Ha: x > 2.58 H0: = 2.58 versus Ha: > 2.58 O Ho: = 2.58 versus Ha: 2.58 b) what is the test statistic? Give your answer to four decimal places. c) What is the P-value for the test? Give your answer to four decimal places. d) What is the appropriate conclusion? O Reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1 O Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is smaller than 0.1 O Fail to reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1 O Reject the claim that the mean number of defects per screen is 2.58 because the P-value is larger than 0.1Explanation / Answer
Appropriate null and alternate hypothesis are
H0: mu = 2.58
H1: mu < 2.58
mean = 1.86
std. dev. = 1.525430685
n = 50
test statistics = (1.86 - 2.58)/1.525430685 = -0.471997848
p-value = 0.319511782
Reject the claim that the mean number of defects per screen is 2.58 because the p-value is larger than 0.1 (option D)
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