Q 10RP. - Elementary Statistics

Question

Hours Actually Worked. In the article "How Hours of Work Affect Occupational Earnings" (Monthly Labor Reriew, Vol. 121), D. Hecker discussed the number of hours actually worked as opposed to the number of hours paid for. The study examines both full-time men and full-time women in $87$ different occupations. According to the article. the mean number of hours (actually) worked by female marketing and advertising managers is $\mu =45$ hours. Assuming a standard deviation of $\sigma =7$ hours, decide whether cach of the following statements is true or false or whether the information is insufficient to decide. Give a reason for each of your answers.

a. For a random sample of $196$ female marketing and advertising managers, chances are roughly $95\%$ that the sample mean number of hours worked will be between $31$ hours and $59$ hours.

b. Approximately $95\%$ of all possible observations of the number of hours worked by female marketing and advertising managers lie between 31 hours and $59$ hours.

c. For a random sample of $196$ female marketing and advertising managers, chances are roughly $95\%$ that the sample mean number of hours worked will be between $44$ hours and $46$ hours.

Step-by-Step Solution

Answer

Part (a) It is around $95\%$ incorrect that the sample's average number of hours worked will be between $31$ and $59$

Part (b) In $95$ percent of all feasible observations, the amount of hours worked by female marketing and advertising managers falls between $31$ and $59$ hours.

Part (c) There is a $95\%$ chance that the sample's average number of hours worked will be between $44$ and $46$

Part (a) Step 1: Given information

Female marketing and advertising managers work an average of $\mu =45$ hours per week (on average).

A $\sigma =7$ hour standard deviation

Part (a) Step 2: Concept

The formula used: Standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Part (a) Step 3: Calculation

The random variable $\overline{x}$ has a mean of ${\mu }_x$ and a standard deviation of $\frac{\sigma }{\sqrt{n}}$ and it is normally distributed. According to the central limit theorem. The standard deviation for sample means is, and ${\mu }_x=45$

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{7}{\sqrt{196}} \\ =\frac{7}{14} \\ =0.5 \end{gathered}$$

Thus 

$$\begin{gathered} P(31\le \overline{x}\le 59)=P\left(\frac{31-45}{0.5}\le \overline{x}\le \frac{59-45}{0.5}\right) \\ =P\left(\frac{-14}{0.5}\le z\le \frac{14}{0.5}\right) \\ =P(-28\le z\le 28) \\ \approx 1 \end{gathered}$$

Thus, it is around $95\%$ incorrect that the sample's average number of hours worked will be between $31$and $59$

Part (b) Step 1: Explanation

The population distribution is not given here. As a result, $95\%$ of all potential observations of the number of hours worked by female marketing and advertising managers fall between $31$ and $59$ hours.

Part (c) Step 1: Calculation

Here,

$$\begin{gathered} P(44\le \overline{x}\le 46)=P\left(\frac{44-45}{0.5}\le \overline{x}\le \frac{46-45}{0.5}\right) \\ =P\left(\frac{-1}{0.5}\le z\le \frac{1}{0.5}\right) \\ =P(-2\le z\le 2) \\ =P(z\le 2)-P(z\le -2) \\ =0.9772-0.0228\left[\begin{array}{l}\text{ From TABLE II: Areas under }\\ \text{ the standard normal curve }\end{array}\right] \\ \$=0.9544\$ \end{gathered}$$

As a result, there is a $95\%$ chance that the sample's average number of hours worked will be between $44$ and $46$