A quality control engineer at a particular lcd screen manufacturer is studying t

Question

A quality control engineer at a particular lcd 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.

What is the test statistic? What is the p-value for the test? Give both answers to four decimal points.

Explanation / Answer

Here we have to test the hypothesis that,

H0 : mu = 2.58 Vs H1 : mu < 2.58

where mu is the population mean number of defects per screen.

mu0 = 2.58

Assume alpha = level of significance = 0.1

Here population standard deviation is unknown so we use t-test for one sample.

This we can done using MINITAB.

Steps :

Enter all the data in MINITAB sheet --> Stat --> Basic Statistics --> 1-Sample t (Test and Confidence Interval) --> Samples in columns : select Column C1 (C1 is that column which has data values) --> Perform hypothesis test --> Hypothesized mean : 2.58 --> options --> Confidence level : 90% --> Alternative : select "<" --> ok --> ok

Output is :

One-Sample T: defects

Test of mu = 2.58 vs < 2.58

90%
Upper
Variable N Mean StDev SE Mean Bound T P
defects 50 1.86000 1.52543 0.21573 2.14025 -3.34 0.001

Test statistic t = -3.34

P-value = 0.001

P-value < alpha

Reject H0 at 10% level of significance.

Conclusion :  The engineer now feels that the mean number of defects per screen may be significantly smaller than 2.58.

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