Question

The following data represents 16 observations of a certain wire thickness for an electrical application: .81 .87 .89 1.04 1.09 1.12 1.29 1.31 1.48 1.49 1.59 1.62 1.65 1.71 1.75 1.85 a) Using Minitab, create a probability plot of the data and determine the type of distribution. b) Compute a point estimate of the mean value of the wire thickness. c) Compute a point estimate of the median of the wire thickness. d) Compute a point estimate of the min/max mean of the wire thickness. e) Compute a point estimate of the 10% trimmed mean of the wire thickness. f) Would you purchase this bran of wire for a mission-critical application? Why?

Answer 1

Tools Window Help Assistant MOOBS @ DESBOGfx -- AI X2 XaRTOO! UM Probability Plot of c1 Probability Plot of C1 Normal - 95% CI Mean 1.348 StDev 0.3418 16 AD 0.389 P-Value 0.342 Percent C14 0.0 0.5 1.0 1.5 2.0 2.5 Ofe À

Minitab - Untitled File Edit Data Calc Stat Graph Editor Tools Window Help Assistant HA1%e4all A1038699 A + YENIX N es Session Probability Plot of C1 Mean of C1 Mean of cl = 1.3475 Median of C1 Median of Cl = 1.395 Minimum of C1 Minimum of Cl = 0.81 Maximum of C1 Maximum of Cl = 1.85 .49 1.59 1.65 1.71 2/61 72x 13 14 1.7 Current Worksheet: Worksheet 1 SEO Type here to search

0 e) Trimmed mean Totey no. of observations. = 16 10.1. trimmed mean. 1680.1 =1.6 w2 Therefore, we semove ist 2 and last 2 otions and compute the mean 0 obsery- 0 trimmed mean = 0.89 +1.04+1.00 +1.12 + 1.29+1.31+ 1.48+1.49 + 1.59 + 1.62 +1.65+1.71 12 0 0 trimmed mean = 12475 1.35667 0 0

In minitab data stored in variable C1.

f. Yes, I will suggest to buy the wire because trimmed mean and mean are almost same.

Even we removed the initial and last measures their is no variation.

The thickness of wire is consistent.

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