The target thickness for silicon wafers used in a certain type of integrated circuit is 245 µm. A sample of 50 wafers is obtained and the thickness of each one is determined, resulting in a sample mean thickness of 246.18 µm and a sample standard deviation of 3.60 µm. Does this data suggest that true average wafer thickness is something other than the target value?1.Parameter of interest: ? = true average wafer thickness2.Null hypothesis: H0: ? 2453.Alternative hypothesis: Ha: ? 2454.Formula for test statistic value: z = x − 245s/ n5.Calculation of test statistic value: z = 246.18 − 2453.60/ 50 = (rounded to two decimal places)6.Determination of P-value: Because the test is two-tailed, P-value = 2(1 − Φ(2.32)) = 0.02047.Conclusion: Using a significance level of 0.01, H0 would not be rejected since 0.0204 > 0.01. At this significance level, there is evidence to conclude that true average thickness differs from the target value.

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The target thickness for silicon wafers used in a certain type of integrated circuit is 245 µm. A sample of 50 wafers is obtained and the thickness of each one is determined, resulting in a sample mean thickness of 246.18 µm and a sample standard deviation of 3.60 µm. Does this data suggest that true average wafer thickness is something other than the target value?

The target thickness for silicon wafers used in a certain type of integrated circuit is 245 µm. A sample of 50 wafers is obtained and the thickness of each one is determined, resulting in a sample mean thickness of 246.18 µm and a sample standard deviation of 3.60 µm. Does this data suggest that true average wafer thickness is something other than the target value?

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

Parameter of interest: ? = true average wafer thickness

Null hypothesis: H₀: ? 245

Alternative hypothesis: H_a: ? 245

Formula for test statistic value:

z =

x − 245

Calculation of test statistic value:

z =

246.18 − 245

3.60/

(rounded to two decimal places)

Determination of P-value: Because the test is two-tailed,

P-value = 2(1 − Φ(2.32)) = 0.0204

Conclusion: Using a significance level of 0.01, H₀ would not be rejected since 0.0204 > 0.01. At this significance level, there is evidence to conclude that true average thickness differs from the target value.

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