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A lathe is set to cut bars of steel into lengths of 6 centimeters. The lathe is considered to be in perfect adjustment if the average length of the bars it cuts is 6 centimeters. A sample of 121 bars is selected randomly and measured. It is determined that the average length of the bars in the sample is 6.08 centimeters with a standard deviation of 0.44 centimeters. Compute the p-value for testing whether the population mean is larger than 6 centimeters.

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Answer:

0.0239

Step-by-step explanation:

[tex]H_0;\mu=\sigma[/tex]

[tex]H_a;\mu>6[/tex]

[tex]\bar{x}=6.08[/tex] = Mean

[tex]s=0.44[/tex] = Standard deviation

[tex]n=121[/tex] = Sample size

Test statistic is given by

[tex]\dfrac{\bar{x}-\mu}{\dfrac{s}{\sqrt{n}}}=\dfrac{6.08-6}{\dfrac{0.44}{\sqrt{121}}}\\ =2[/tex]

From T table we know it is a right tailed test.

Degree of freedom = [tex]n-1=121-1=120[/tex]

[tex]P(t>2)=1-P(t<2)\\\Rightarrow P(t>2)=1-0.9761\\\Rightarrow P(t>2)=0.0239[/tex]

Hence, p-value is 0.0239.

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