Question

The number of television viewing hours per household and the prime viewing times are two factors that affect television advertising income. A random sample of 25 households in a particular viewing area produced the following estimates of viewing hours per household: $$ \begin{array}{rrrrr} 3.0 & 6.0 & 7.5 & 15.0 & 12.0 \\ 6.5 & 8.0 & 4.0 & 5.5 & 6.0 \\ 5.0 & 12.0 & 1.0 & 3.5 & 3.0 \\ 7.5 & 5.0 & 10.0 & 8.0 & 3.5 \\ 9.0 & 2.0 & 6.5 & 1.0 & 5.0 \end{array} $$ a. Scan the data and use the range to find an approximate value for \(s\). Use this value to check your calculations in part \(\mathrm{b}\). b. Calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\). Compare \(s\) with the approximate value obtained in part a. c. Find the percentage of the viewing hours per household that falls into the interval \(\bar{x} \pm 2 s\). Compare with the corresponding percentage given by the Empirical Rule.

Short answer

Answer: Approximately 84% of television viewing hours per household falls into the interval x̄ ± 2s. This is close to the 95% expected by the Empirical Rule.

Step-by-step solution

Calculate the range of the data

To find the range, first locate the highest and lowest values in the dataset and then subtract the lowest value from the highest value. Max: 15.0 Min: 1.0 Range: 15.0 - 1.0 = 14.0

Approximate the standard deviation using the range

To find an approximate value for standard deviation (s), divide the range by 4. s ≈ (14.0) / 4 = 3.5

Calculate the sample mean

To find the sample mean (x̄), sum the given data points (viewing hours) and then divide by the total number of observations (25). x̄ = (3+6+7.5+15+12+6.5+8+4+5.5+6+5+12+1+3.5+3+7.5+5+10+8+3.5+9+2+6.5+1+5)/25 = 132.5/25 = 5.3

Calculate the sample standard deviation

To calculate the sample standard deviation (s), first find the squared differences between each observation and the sample mean(x̄), then find the sum of those squared differences. Next, divide the sum by (n-1) to yield the squared variance, where n is the number of observations (25 in this case). Lastly, find the square root of the squared variance to obtain the sample standard deviation. 1. Calculate squared differences: (3-5.3)^2 = 5.29 (6-5.3)^2 = 0.49 … (5-5.3)^2 = 0.09 2. Calculate the sum of squared differences: 5.29 + 0.49 + ... + 0.09 = 159.21 3. Calculate squared variance: Squared Variance (s^2) = (159.21) / (25-1) = 159.21 / 24 = 6.63375 4. Calculate the sample standard deviation (s) as the square root of the variance: s = sqrt(6.63375) = 2.57 Comparing the approximated value of s (3.5) from step 2, we can see that the actual value (2.57) is smaller than the approximated value.

Find the interval x̄ ± 2s and percentage of viewing hours

To find the interval, first determine x̄ - 2s and x̄ + 2s. x̄ - 2s = 5.3 - 2(2.57) = 0.16 x̄ + 2s = 5.3 + 2(2.57) = 10.44 The interval is [0.16, 10.44]. Now, count the number of households whose viewing hours fall into this interval and calculate the percentage: In the interval: 21 households Total households: 25 Percentage of viewing hours in the interval = (21/25)*100% = 84% Now, let's compare it with the Empirical Rule, which states that approximately 68% of the data should fall within x̄ ± s; approximately 95% should fall within x̄ ± 2s; and approximately 99.7% should fall within x̄ ± 3s. In our case, 84% of the data falls within x̄ ± 2s, which is close to the 95% expected by the Empirical Rule.

Key concepts

Sample Mean

The "sample mean" is a measure of central tendency for a dataset. It is calculated by summing all the observations and then dividing the total by the number of observations. This statistic provides an average value that represents the dataset as a whole. By looking at the sample mean, you get an idea of where the center point of your data lies.

For students handling datasets, it's important to recognize the sample mean's role in statistical analysis. In our example of television viewing hours, the sample mean is calculated by adding up all the individual viewing hours and dividing by the number of households, which in this case is 25. Hence, the computation is:

\[ \bar{x} = \frac{3+6+7.5+15+12+6.5+8+4+5.5+6+5+12+1+3.5+3+7.5+5+10+8+3.5+9+2+6.5+1+5}{25} = 5.3 \]

Here, the sample mean is approximately 5.3 hours. This value gives a snapshot of the average TV viewing time per household in our sample.

Sample Standard Deviation

"Sample standard deviation" is a useful statistic that describes how much each data point in a sample tends to deviate from the sample mean. It's a measure of the dispersion or spread within a dataset. When calculating it, the first step is to find the variance, which is the average of the squared differences from the mean. Finally, the square root of the variance is taken to obtain the standard deviation.

In a nutshell, a higher standard deviation indicates more variability among the data points, while a lower standard deviation suggests that the data points are closer to the mean.

For our example, calculating the standard deviation involves several steps:

• Compute the squared differences from the sample mean, \(\bar{x}\), for each observation.
• Calculate the sum of these squared differences, which was found to be 159.21.
• Divide this sum by the number of observations minus one (n-1), where n=25, to find the variance: \( \text{Variance} = \frac{159.21}{24} = 6.63375 \).
• Take the square root of the variance to obtain the sample standard deviation: \( s = \sqrt{6.63375} \approx 2.57 \).

This value, 2.57, tells us the extent of variation in household TV viewing hours around the mean.

Empirical Rule

The "Empirical Rule" is a statistical rule that applies to datasets with a normal distribution. It can provide excellent insight into how data points are dispersed around the mean in a bell-shaped curve. This rule is sometimes called the 68-95-99.7 rule because approximately 68% of observations fall within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

This rule allows a quick estimate of the spread of data in terms of standard deviations away from the mean, offering an intuitive grasp of how concentrated the data is.

In the TV viewing example provided, this rule helps us determine the concentration of data around the mean:\[ \bar{x} \pm 2s \]. First, we calculate the interval, \[ \bar{x} - 2s \] to \[ \bar{x} + 2s \], which is \[0.16\] to \[10.44\]. Subsequently, we find that 84% of the households fall within this range.

Although this is slightly lower than the expected 95% indicated by the Empirical Rule for a normal distribution, it still provides valuable insights. It suggests that most households' viewing hours are within these bounds, giving advertising agencies a clearer picture of typical viewing behaviors.