Question

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.

Answer 1

0.0239

Explanation:

`H_0;\mu=\sigma`

`H_a;\mu>6`

`\bar{x}=6.08` = Mean

`s=0.44` = Standard deviation

`n=121` = Sample size

Test statistic is given by

`\frac{\bar{x}-\mu}{(s)/(√(n))}=(6.08-6)/((0.44)/(√(121)))\\ =2`

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

Degree of freedom =
`n-1=121-1=120`

`P(t>2)=1-P(t<2)\\\Rightarrow P(t>2)=1-0.9761\\\Rightarrow P(t>2)=0.0239`

Hence, p-value is 0.0239.