Mean and Standard Error

• The mean of the sampling distribution is 180 hours or, $\mu_{\bar{x}}= 180$.
• The question is asking about the sample mean – it is not asking about any one individual Canadian 18 years of age or older
• The question is a sampling distribution question so we need to use the standard error formula.
• The standard error is $\sigma_{\bar{x}}= 11.25$ hours: $$\sigma_{\bar{x}}= \frac{\sigma}{\sqrt{n}} = \frac{90}{\sqrt{64}} = 11.25 \text{ hours}$$

Graph of Area in Question

As with any other normal distribution problem, we start by drawing a distribution of the sampling distribution and label it appropriately with the information in the problem:

Excel Formula

To solve for the probability of the sample mean ($\bar{x}$) being greater than 200 hours, we use the following Excel formula: $$=1-\text{NORM.DIST}(200,180,11.25,1) = 0.0377$$ .

Conclusion

There is a 3.77% chance that the sample average will be larger than 200 hours.

Mean and Standard Error

• The mean is the same as the previous example: $\mu_{\bar{x}}= 180$.
• Again, the question is asking about the sample mean and not about a particular individual so we are working with a sampling distribution question and need to use a standard error.
• The standard error is still $\sigma_{\bar{x}}= 11.25$ hours: $$\sigma_{\bar{x}}= \frac{\sigma}{\sqrt{n}} = \frac{90}{\sqrt{64}} = 11.25 \text{ hours}$$

Graph of Area in Question

As with any other normal distribution problem, we start by drawing a distribution of the sampling distribution and label it appropriately with the information in the problem:

Excel Formula

To solve for the probability of the sample mean ($\bar{x}$) being less than 150 hours, we use the following Excel formula: $$=\text{NORM.DIST}(150,180,11.25,1) = 0.0038$$ .

Conclusion

There is a 0.38% chance that the sample average will be less than 150 hours.

This question is asking the likelihood that the sample mean will be within 10 hours of the population – in other words, the probability that the sample mean will fall between -10 hours below the population mean to +10 hours above the population mean. This is illustrated as follows in Figure 46.3:

Here we only need to find the area on one side of the sampling distribution and then we can multiply it by 2 because we are dealing with a symmetrical distance of 10 hours above and below the population mean.

Using our sampling formula, we will determine the area between the center of the sampling distribution to a sample mean of 190 hours. Thus the numerator is +10 (i.e., the difference between the sample mean of 190 and the population mean of 180) and the corresponding z-score will be:

$$Z = \frac{\bar{x}-\mu}{\sigma_{\bar{x}}} = \frac{10}{11.25} = 0.8889$$

Thus, we round up our z-score to 0.89 and look up the corresponding area on the cumulative standardized normal table. A z-score of 0.89 corresponds to an area of 0.8106.. This is the area that falls less than a sample mean of 190 hours. Thus, the area from the center of our sampling distribution to the sample mean of 190 hours is: $0.8106 - 0.5 = 0.3106$.

Next we must multiply this area by 2 because the question is asking about the likelihood of the sample mean being within 10 hours (i.e., on either side) of the population mean: $0.3106 \times 2 = 0.6212$

Thus the likelihood of taking a sample of 64 adults and finding that they spend an average of 10 hours per year within the true population mean of 180 hours using the internet is 62.12% (quite likely!).

This question is asking us to determine the 95th percentile of the sample means with respect to the sampling distribution. In other words, what is value of the sample mean whereby 95% of all sample means fall less than this value (5% fall above this value).

We can illustrate this as follows in Figure 46.4:

Using our sampling formula, we know all everything except a z-score and the sample mean:

$$Z = \frac{\bar{x}-\mu}{\sigma_{\bar{x}}} = \frac{\bar{x}-180}{11.25}$$

But recall, if we have an Area we can determine its corresponding z-score with the normal table. Thus, an area of 0.95 corresponds to the z-score of +1.645. We now have the following:

$$\frac{\bar{x}-180}{11.25} = 1.645$$

Solving for the sample mean we get: $\bar{x} = 180 +1.645\cdot 11.25=198.51$ hours hours (rounded to 2 decimal places). Thus 95% of all sample means will fall below 198.51 hours per year of internet use.