Question

How many tissues should the Kimberly Clark Corporation package of Kleenex contain? Researchers determined that 60 tissues is the mean number of tissues used during a cold. Suppose a random sample of 100 Kleenex users yielded the following data on the number of tissues used during a cold: X= 55, S = 20. Using the sample information provided, calculate the value of the test statistic.


A)z = (55 - 60) / 20


B)z = (55 - 60) / (20/100)


C)z = (55 - 60) / (20/1002)


D)z = (55 - 60) / (20/10)

Answer 1

To calculate the value of the test statistic, we need to use the formula for the z-test statistic, which is given as:

\[
z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}
\]

Where:
- \(\bar{X}\) is the sample mean (which in this case is 55).
- \(\mu\) is the population mean (which in this case is 60).
- \(\sigma\) is the sample standard deviation (which is given as 20).
- \(n\) is the sample size (which is 100).

Now, substituting the values into the formula:

1. The difference between the sample mean and the population mean is:
\[
\bar{X} - \mu = 55 - 60 = -5
\]

2. The standard error of the mean is calculated as:
\[
\sigma / \sqrt{n} = 20 / \sqrt{100} = 20 / 10 = 2
\]

3. Now, we can calculate the z-value:
\[
z = \frac{-5}{2} = -2.5
\]

Now let's evaluate the options provided:

A) \(z = \frac{(55 - 60)}{20}\)
B) \(z = \frac{(55 - 60)}{(20/100)}\)
C) \(z = \frac{(55 - 60)}{(20/100^2)}\)
D) \(z = \frac{(55 - 60)}{(20/10)}\)

- Option A: Incorrect, since it's not dividing by the standard error.
- Option B: Incorrect, it should be the standard error \(20/\sqrt{100}\), not \(20/100\).
- Option C: Incorrect, the denominator should not have \(100^2\).
- Option D: Correct, as it uses the right standard error \((20/10)\).

Based on the analysis and calculations, the correct expression used for the z-value is represented in **Option D**, \(z = \frac{(55 - 60)}{(20/10)}\), even though the actual value of \(z\) after computation is not required. The method confirms your approach is valid.