Question

Suppose a sales department gets an average of 20.1 complaints per week with a population standard deviation of 1.4 complaints. Based on the complaints, the manager suspects cultural diversity training will help. Everyone takes the training. The manager notices that after the training, the complaints drop to 18.9 per week (based on five weeks, or n=5). What's the absolute value of your calculated value? Round your answer to two decimal places.

Answer 1

`\mid z \mid = 1.92`

Explanation:

The null hypothesis is that the average number of complaints per week is 20.1

If H₀ = Null hypothesis

H₀: μ = 20.1

After the training, the complaints drop to 18.9 < 20.1. This means that the alternative hypothesis,
`H_(a) : \mu < 20.1`

`\bar{x} = 18.9`

To know if the alternative hypothesis is true, we need to calculate the absolute z value using the test statistic

number of days, n = 5

Standard deviation,
`\sigma = 1.4`

`z = \frac{\bar{x} - \mu}{(\sigma)/(√(n) ) }`

`z = (18.9 - 20.1)/((1.4)/(√(5) ) )`

z = -1.92

`\mid z \mid = 1.92`

Answer 2

The absolute value is
`|z| = 1.92`

Explanation:

From the question we are told that

The mean is
`\mu = 20.1`

The standard deviation is
`\sigma = 1.4`

For the null hypothesis
`H_0`

The mean remains
`\mu = 20.1`

For the alternative hypothesis
`H_a`

The mean is
`\mu < 20`

This mean that the claim drops

The test statistic(the calculated value ) (z) is mathematically obtained with the following formula

`z = (\= x -\mu)/((\sigma)/(√(n) ) )`

Where
`\= x` is the mean for the the alternative hypothesis

substituting values

`z = (18.9 - 20.1)/((1.4)/(√(5) ) )`

`= -1.92`

the absolute value of the calculated value is

`|z| = 1.92`