Question

In preparation for upcoming wage negotiations with the union, the managers for the Bevel Hardware Company want to establish the time required to assemble a kitchen cabinet. A first line supervisor believes that the job should take 3030 minutes on average to complete. A random sample of 150150 cabinets has an average assembly time of 5555 minutes with a standard deviation of 100100 minutes. Is there overwhelming evidence to contradict the first line supervisor’s belief at a 0.050.05 significance level?

Step 1 of 3 :  

Find the value of the test statistic. Round your answer to three decimal places, if necessar

Answer 1

Given: Population mean (µ) = 30

Sample size(n) = 150

Sample mean = xbar = 55
Sample standard deviation = S = 100

The e Null hypothesis is, H0: µ = 30

And the alternative hypothesis is,

Here, we have the sample standard deviation. So, we need to use the T-test for testing.

Now, let's find the test statistic.

That is, test statistic T = 3.062

Degrees of freedom = n -1 = 150 - 1 = 149

Now, let's find the P-value.
For finding the P-value, we can use a technology like excel/Ti84 calculator or T table.
The following excel command is used to find the P-value.

= T.DIST.2T(Test statistic, Degrees of freedom)
= T.DIST.2T(3.062 , 149 )

You will get, P-value = 0.0026

We are given: Level of significance α= 0.05

Following is the decision rule for making the decision.
If, P-value > α, then we fail to reject the null hypothesis.
If, P-value <= α, then we reject the null hypothesis.

α = 0.05
And P-value = 0.0026
That is P-value < α.
Hence, we reject the null hypothesis.
That is, there is sufficient evidence to contradict the first line supervisor's belief at a 0.05 significance level.