OPTIMA Statistics

No! The checks you sampled include a tip. They will have a distribution that is centered above the true average price of a lunch without tip.

Both samples have the same mean, \(\overline x = 20\text{.}\) So there would be no difference in their point estimates. However, the variances in the samples are very different. Sample (a) suggests that the real population mean, while close to 20, could be quite different because the sample has such a large range. On the other hand, sample (b) suggests that the real population mean should be very close to 20 since the sample is not spread out very much.

To find the critical values for these confidence levels, we must find the z-score that separates the middle \((1-\alpha)\) percent of the data from the top and bottom \(\alpha/2\) percent of the data.

1. For the 95% confidence level, there is 0.9500 in the middle of the normal distribution, and half of the remaining 0.0500 in each of the tails. That means that there is 0.0250 in the upper tail and 0.0250 in the lower tail as shown below.

   

   Looking up the probability 0.0250 in the body of negative part of the standard normal distribution table, we find the z-score \(-1.96\text{.}\) Because the standard normal distribution is symmetric, we know that if we looked up the probability \(1-0.0250 = 0.9750\) in the body of the positive part of the table, we would find the positive version of this z-score. Thus the critical values are \(-z_{\alpha/2} = -1.96\) and \(z_{\alpha/2} = 1.96\text{.}\)

2. For the 98% confidence level, the process is similar. There is 0.9800 in the middle of the normal distribution leaving 0.0200 in the tails. Dividing this in half, there is 0.0100 in the upper tail and 0.0100 in the lower tail, as shown below.

   

   Again looking up the lower tail probability 0.0100 we find the closest probability in the table is 0.0099 which goes with a z-score of \(-2.33\text{.}\) Therefore, the critical values are \(-z_{\alpha/2} = -2.33\) and \(z_{\alpha/2} = 2.33\text{.}\)

The margin of error is \(\sigma_x\) times \(z_{\alpha/2}\text{.}\) Based on our work in Example 4.1.22, this gives the following values.

1. For a 95% confidence interval, \(z_{\alpha/2} = \pm 1.96\) so that the margin of error is:

   \begin{equation*} \pm 1.96 \times 4.35 = \pm 8.526\text{.} \end{equation*}

2. For a 98% confidence interval, \(z_{\alpha/2} = \pm 2.33\) so that the margin of error is:

   \begin{equation*} \pm 2.33 \times 4.35 = \pm 10.1355\text{.} \end{equation*}

1. 95% Upper Confidence Bound.

   To get a 95% upper confidence bound, we need a critical value \(z_{0.05}\) that separates the bottom 95% of the data from the top 5%. Looking up \(0.9500\) in the standard normal distribution table, we find \(0.9495\) goes with a z-score of \(1.64\) and \(0.9505\) goes with \(1.65\text{.}\) So the z-score we want is half-way between \(1.64\) and \(1.65\text{,}\) which gives \(z_\alpha = 1.645\) (Note: we only average this way because 0.95 is exactly half way between \(0.9495\) and \(0.9505\)). Then, computing the upper confidence bound, we get:

   \begin{equation*} 10 + 1.645 \times 1.5 = 10 + 2.4675 = 12.4675\text{.} \end{equation*}

2. 98% Lower Confidence Bound.

   In this case we want 2% in the bottom tail and 98% in the top tail. Looking up \(0.0200\) in the standard normal distribution table we find a z-score of \(-2.05\text{.}\) Note that the critical value is negative, so when we multiply by the standard error and add it to the point estimate, we will actually subtract. This computation yields:

   \begin{equation*} 10 - 2.05 \times 1.5 = 10 - 3.075 = 6.925\text{.} \end{equation*}