Question

In: Statistics and Probability

It is known that the mean income for all assembly-line workers with less than 5 years...

It is known that the mean income for all assembly-line workers with less than 5 years of experience in a large compay is $300 per week. A representative of a women’s group believes that the female employees are being underpaid. A random sample of sixteen female employees yields an average weekly income of $270. Conduct a statistical test that will give you evidence about whether the female employees have a mean income of less than $300 per week. It is assumed that the weekly incomes for all the female employees follows a Normal Distribution with standard deviation σ = $36.

• What does the Central Limit Theorem say under the assumption that the Null is true?

•Draw the Distribution and mark off your sample mean.

• Compute the Test Statistic: z = x¯−µ / √σ /n

• Calculate the p-value

• Do you Reject the H0 or Fail to Reject the Ho?

• Write out a statement describing your conclusion.

Solutions

Expert Solution

• What does the Central Limit Theorem say under the assumption that the Null is true?

The central limit theorem say under the assumption that the null is true, that the sampling distribution of the sample means follows an approximate normal distribution although the original distribution is normal or not normal.

•Draw the Distribution and mark off your sample mean.

Required normal curve is given as below:

• Compute the Test Statistic:

H0: µ = 300 versus Ha: µ < 300

We are given Xbar = 270, σ = 36, n = 16, µ = 300

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (270 – 300)/[36/sqrt(16)]

Z = -30/[36/4]

Z = -30/9

Z = -3.33333

Test statistic = -3.33333

• Calculate the p-value

P-value = 0.0004

(by using z-table)

• Do you Reject the H0 or Fail to Reject the Ho?

Here,

P-value = 0.0004

α = 0.05

P-value < α

So, we reject the H0

• Write out a statement describing your conclusion.

There is sufficient evidence to conclude that the female employees have a mean income of less than $300 per week.


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