Question

Managers at an automobile manufacturing plant would like to examine the mean completion time of an assembly line operation. The past data indicate that the mean completion time is 43 minutes, but the managers have reason to believe that this value has decreased. The managers plan to perform a statistical test of that belief. After choosing a random sample of 9 assembly line completion times, the managers compute the sample mean completion time to be 40 minutes and the standard deviation to be 3 minutes. (You may use answers to other questions in this test.) What are the value of the test statistic and the P-value?

Answer 1

The value of t test statistics is -3.

The P-value is 0.0089.

Explanation:

We are given that the past data indicate that the mean completion time is 43 minutes, but the managers have reason to believe that this value has decreased.

After choosing a random sample of 9 assembly line completion times, the managers compute the sample mean completion time to be 40 minutes and the standard deviation to be 3 minutes.

Let
`\mu` = mean mean completion time.

So, Null Hypothesis,
`H_0` :
`\mu \geq` 43 minutes {means that the mean completion time of an assembly line operation has increased or remains same}

Alternate Hypothesis,
`H_A` :
`\mu` < 43 minutes {means that the mean completion time of an assembly line operation has decreased}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. =
`(\bar X-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean completion time = 40 minutes

s = sample standard deviation = 3 minutes

n = sample of assembly line completion times = 9

So, test statistics =
`(40-43)/((3)/(√(9) ) )` ~
`t_8`

= -3

The value of t test statistics is -3.

Now, the P-value of the test statistics is given by the following formula;

P-value = P(
`t_8` < -3) = 0.0089 or 0.89%