The data below contain the number of defects observed on each of 50 lcd screens

Question

The data below contain the number of defects observed on each of 50 lcd screens made by a certain manufacturer:

2

2 2 0 4 4 5 4 3 5 1 4 2 1 2 0 2 0 4 4 1 6 2 0 5 2 0 0 1 1 3 2 4 1 3 3 0 2 2 3 2 4 2 1 1 1 1 2 1 0

2

Construct a 95% confidence interval for the mean number of defects per screen for all screens produced by this manufacturer a) What is the lower limit on the interval? Give your answer to three decimal places b) What is the upper init n Lhe interval? Give your answer to three decirnal places. c) Based on the interval above, would you believe that the mean number of defects per screen tor all screens from this manufacturer is 1.78? No because 1.78 is inside the interval. Yes because 1./8 is not inside the interval No becase 1.78 is rnot inside the interval. Yes because I .78 is inside the interval. d) If we are ver certain that the true standard deviation of the number of defects per screen 5 below 2, what samples ze would e reure 50 that of defects per screen is at most 0.34? Make sure you enter a whole nurnber below. e 1 tn of the % conf denceinte a for the mean num

Explanation / Answer

Mean = 2.14

Standard deviation = 1.56

At 95% confidence interval the critical value is 1.96

Confidence interval = mean + /- Z * SD/sqrt (n)

= 2.14 + / - 1.96 * 1.56/sqrt(50)

= 2.14 + /- 0.432

= 1.708, 2.572

A) Lower limit of the interval is 1.708

B) Upper limit of the interval is 2.572

C) Option-D is the correct answer.

D) 0.34 = 2.14 + 1.96 * 1.56/sqrt(n)

Or, -1.8 = 1.96 * 1.56/Sqrt (n)

Or, sqrt (n) = (1.96 * 1.56)/(-1.8)

Or, sqrt(n) = 1.699

Or, n = 2.8866 = 3

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