Sampling Distribution of Means (Continued)

• Example 8.6: An electric firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 800 hours and a standard deviation of 40 hours.
• Find the probability that a random sample of 16 bulbs will have an average life of less than 775 hours.
• Solution:
  • Even though $16 <30$, the central limit theorem can be used because it is stated that the population distribution is approximately normal.
  • The sampling distribution of $\bar{X}$ will be approximately normal, with
    $$\mu_{\bar{X}}=800,~\sigma_{\bar{X}}=40/\sqrt{16}=10$$
    $$\bar{x}=75 \Rightarrow z=\frac{775-800}{10}=-2.5$$
    $$P(\bar{X}< 775)=P(Z < -2.5)=0.0062$$

  Figure 7.11: Area for Example 8.6.

  

• Example 8.7: A engineer conjectures that the population mean of a certain component parts is 5.0 millimeters. An experiment is conducted in which 100 parts produced by the process are selected randomly and the diameter measured on each.
• It is known that the population standard deviation $\sigma = 0.1$. The experiment indicates a sample average diameter $\bar{X} =5.027$ millimeters.
• Does this sample information appear to support or refute the engineer's conjecture?
• Solution:

$$P\left[ \vert(\bar{X}-5)\vert \geq 0.027 \right]$$

$$=P[(\bar{X}-5) \geq 0.027]+P[(\bar{X}-5) \leq -0.027]$$

$$=2P(Z\geq2.7)=2*0.0035=0.007$$

Strongly refutes the conjecture!

Figure 7.12: Area for Example 8.7.

• Sometimes we are interested in comparing two populations (i.e., one manufacturing process better than the other).
• Suppose we have two populations, the first with $\mu_1$ and $\sigma_1$ and the second with $\mu_2$ and $\sigma_2$.
• Let the statistic $\bar{X_1}$ represent the sample mean selected from the first population and the statistic $\bar{X_2}$ represent the sample mean selected from the second population.
• How about the sampling distribution
• Solution: Using Theorem 7.11, $\bar{X_1}-\bar{X_2}$ is approximately normally distributed with mean
  $$\mu_{\bar{X_1}-\bar{X_2}}=\mu_{\bar{X_1}}-\mu_{\bar{X_2}}=\mu_1-\mu_2$$
  and variance
  $$\sigma_{\bar{X_1}-\bar{X_2}}^2=\sigma_{\bar{X_1}}^2+\sigma_{\bar{X_2}}^2=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$$

• Theorem 8.3:
  

• Example 8.8: Two independent experiments are being run in which two different types of paints are compared.
• Eighteen specimens are painted using type $A$ and the drying time in hours is recorded on each. The same is done with type $B$.
• The population standard deviations are both known to be 1.0. Assuming that the mean drying time is equal for the two types of paint,
• find $P(\bar{X_A}-\bar{X_B} > 1.0)$ where $\bar{X_A}$ and $\bar{X_B}$ are average drying times for samples of size $n_A=n_B=18$.
• Solution:

$$\mu_{\bar{X_A}-\bar{X_B}}=\mu_A-\mu_B=0$$

$$\sigma_{\bar{X_A}-\bar{X_B}}^2=\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}$$

$$z=\frac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}}=\frac{1-0}{\sqrt{1/9}}=3.0$$

$$P(Z>3.0)=1-P(Z<3.0)$$

$$=1-0.9987=0.0013$$

Figure 7.13: Area for Example 8.8.

Low probability. Assumption?

• Example 8.9: The television picture tubes of manufacturer $A$ have a mean lifetime of 6.5 years and a standard deviation of 0.9 year, while those of manufacturer $B$ have a mean lifetime of 6.0 years and a standard deviation of 0.8 year.
• What is the probability that a random sample of 36 tubes from manufacturer $A$ will have a mean lifetime that is at least 1 year more than the mean lifetime of a sample of 49 tubes from manufacturer $B$?
  

  Table 7.2: Data for Example 8.9.

  Population 1	Population 2
  $\mu_1$= 6.5	$\mu_2$= 6.0
  $\sigma_1$= 0.9	$\sigma_2$= 0.8
  $n_1$= 36	$n_2=49$

  
  
• $P(\bar{X_1}-\bar{X_2} \geq 1.0) =?$

Solution: Since both $n_1$ and $n_2$ is greater than 30, the sampling distribution of $\bar{X_1}-\bar{X_2}$ will be approximately normal.

$$\mu_{\bar{X_1}-\bar{X_2}}=\mu_1-\mu_2=6.5-6.0=0.5$$

$$\sigma_{\bar{X_1}-\bar{X_2}}^2=\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}$$

$$=\frac{0.9^2}{36}+\frac{0.8^2}{49}=0.0356$$

$$z=\frac{1-0.5}{\sqrt{0.0356}}=2.65$$

$$P(\bar{X_1}-\bar{X_2} \geq 1.0)=P(Z> 2.65)$$

$$=1-P(Z<2.65)=1-0.9960$$

$$=0.004$$

Low probability value.

Figure 7.14: Area for Example 8.9.