91Ó°ÊÓ

A marketing research experiment was conducted to study the relationship between the length of time necessary for a buyer to reach a decision and the number of alternative package designs of a product presented. Brand names were eliminated from the packages to reduce the effects of brand preferences. The buyers made their selections using the manufacturer's product descriptions on the packages as the only buying guide. The length of time necessary to reach a decision was recorded for 15 participants in the marketing research study. $$ \begin{array}{l|l|l|l} \begin{array}{l} \text { Length of Decision } \\ \text { Time, } y(\mathrm{sec}) \end{array} & 5,8,8,7,9 & 7,9,8,9,10 & 10,11,10,12,9 \\ \hline \text { Number of } & & & \\ \text { Alternatives, } x & 2 & 3 & 4 \end{array} $$ a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Calculate \(s^{2}\). d. Do the data present sufficient evidence to indicate that the length of decision time is linearly related to the number of alternative package designs? (Test at the \(\alpha=.05\) level of significance.) e. Find the approximate \(p\) -value for the test and interpret its value. f. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. g. Estimate the average length of time necessary to reach a decision when three alternatives are presented, using a \(95 \%\) confidence interval.

Step by step solution

01

Calculate the sums and average values

To find the least-squares line, we first need to calculate the sums and average values of x, y, xy, and \(x^2\). Create a table to organize the data and calculate these values: $$ \begin{array}{c|c|c|c|c} \text {Alternatives }(x) & \text {Decision Time }(y) & xy & x^2 & y^2 \\ \hline 2 & 5 & 10 & 4 & 25 \\ 2 & 8 & 16 & 4 & 64 \\ 2 & 8 & 16 & 4 & 64 \\ 2 & 7 & 14 & 4 & 49 \\ 2 & 9 & 18 & 4 & 81 \\ \hline 3 & 7 & 21 & 9 & 49 \\ 3 & 9 & 27 & 9 & 81 \\ 3 & 8 & 24 & 9 & 64 \\ 3 & 9 & 27 & 9 & 81 \\ 3 & 10 & 30 & 9 & 100\\ \hline 4 & 10 & 40 & 16 & 100\\ 4 & 11 & 44 & 16 & 121\\ 4 & 10 & 40 & 16 & 100\\ 4 & 12 & 48 & 16 & 144\\ 4 & 9 & 36 & 16 & 81 \\ \hline \sum & \sum & \sum_{xy} \ & \sum_{x^2} & \sum_{y^2} \\ \end{array} $$ With these values, we can calculate the sums and averages: \(\sum x = 45, \sum y = 135, \sum xy = 337, \sum x^2 = 153, \sum y^2 = 1389\) \(N = 15\) (number of participants) \(\overline{x}=\frac{1}{15} \sum x = \frac{45}{15} = 3, \overline{y} = \frac{1}{15}\sum y = \frac{135}{15} = 9\).

02

Find the least-squares line

Now, using the values we found in step 1, we can calculate the slope (\(b_1\)) and the y-intercept (\(b_0\)) of the least-squares line using the following formulas: \(b_1 = \frac{N\cdot\sum(xy)-\sum x \cdot \sum y}{N\cdot \sum x^2 - (\sum x)^2} = \frac{15\cdot 337 - 45\cdot135}{15\cdot153 - 45^2} = 2.40\) \(b_0 = \overline{y} - b_1\cdot\overline{x} = 9 - 2.40\cdot3 = 2.20\) So, the least-squares line is \(y = 2.20 + 2.40x\).

03

Calculate the variance \(s^2\)

The variance (\(s^2\)) can be calculated using the following formula: \(s^2 = \frac{\sum y^2 - b_0\cdot \sum y - b_1\cdot \sum xy}{N-2} = \frac{1389 - 2.20\cdot 135 - 2.40\cdot 337}{15-2} = 1.635\)

04

Compute the test statistic, critical value, and decision rule

Since we are testing at the \(\alpha = .05\) level of significance, we can compute the test statistic, critical value, and decision rule for a two-tailed test. We will use the t-test statistic, given by: \(t = \frac{b_1}{s\cdot\sqrt{\frac{1}{\sum{x^2} - \frac{(\sum x)^2}{N}}}}\) where s is the standard deviation, which can be obtained by taking the square root of the variance (\(s^2\) found in step 3). \(s = \sqrt{1.635} \approx 1.27861\) Hence the test statistic: \(t = \frac{2.40}{1.27861\cdot\sqrt{\frac{1}{153 - \frac{45^2}{15}}}} \approx 2.391\) Degrees of freedom = \(N-2 = 15-2 = 13\). Using the t-distribution table, the critical value for \(α/2=0.025\) and 13 degrees of freedom is about \(2.160\). The decision rule is that if the absolute value of t is greater than the critical value, we reject the null hypothesis. Since our calculated test statistic \(|t| = 2.391 > 2.160\), we reject the null hypothesis.

05

Find the approximate p-value

To find the approximate p-value for our test, we can look up the probability of obtaining a value as extreme or more extreme than our test statistic in the t-distribution table. For our calculated test statistic \(t = 2.391\) and 13 degrees of freedom, we find that the probability is between \(0.02\) and \(0.05\). Hence, the approximate p-value is between \(0.02\) and \(0.05\). Since the p-value is less than \(\alpha = .05\), it provides further evidence to reject the null hypothesis.

06

Examine the diagnostic plots

In the absence of diagnostic plots, we will not be able to check the validity of the regression assumptions.

07

Estimate the average length of decision time for three alternatives

To find the average length of decision time when three alternatives are presented, substitute \(x = 3\) into the least-squares line equation: \(y = 2.20 + 2.40 \cdot 3 = 9.40\) seconds Now, to calculate the \(95\%\) confidence interval, we need the standard error of the estimate \(s_e\): \(s_e = s\cdot\sqrt{\frac{1}{N}+\frac{(x-\overline{x})^2}{\sum(x^2)-\frac{(\sum{x})^2}{N}}} = 1.27861\cdot\sqrt{\frac{1}{15}+\frac{(3-3)^2}{153-\frac{(45)^2}{15}}} = 0.33025\) With 13 degrees of freedom, and using the t-distribution table at \(0.025\) (two-tailed) we have \(t = 2.160\). Confidence interval: \(y \pm t\cdot s_e\) \(9.40 \pm 2.160\cdot0.33025\) \(9.40 \pm 0.71294\) \(8.69 \le y \le 10.11\) Thus, we can estimate that when three alternatives are presented, the average length of time necessary to reach a decision is between \(8.69\) seconds and \(10.11\) seconds with a \(95\%\) confidence level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Least Squares Line

When dealing with regression analysis, one of the most fundamental concepts is the 'least squares line'. This is essentially the best-fitting straight line through a set of points in a scatter plot, representing the relationship between two variables. In the context of the marketing research experiment, we're looking at the time taken to make a decision (dependent variable, y) in relation to the number of package designs (independent variable, x).

The calculation of the least squares line involves finding the line that minimizes the sum of the squares of the vertical distances (residuals) between the observed values and the line's predicted values. The equation for the least squares line is given by y = b₀ + b₁x, where b₀ is the y-intercept, and b₁ is the slope.

• The slope gives us the rate at which the decision time changes for each additional package design.
• The y-intercept indicates the decision time when no package designs are presented.

While the least squares line provides valuable insight into the relationship, the exercise should also include a visual check by plotting the points and the line, ensuring that the line fits well with the observed data.

Statistical Significance

Determining whether the results of a study are due to chance or an actual relationship between variables is pivotal. This is where 'statistical significance' comes into play. In regression, we often use a hypothesis test to decide whether the slope of the regression line differs significantly from zero, which would mean there is no relationship.

In our marketing research example, we calculate a test statistic to compare against a critical value. If our test statistic is greater than the critical value, or if the p-value is less than our chosen significance level (commonly set at 0.05), we have evidence to reject the null hypothesis, which typically posits that there is no relationship. This is what happened in our case: the finding suggests that there is a statistically significant relationship between the number of package designs and decision time. For a student conducting similar analysis, ensuring the correct formulation of hypotheses and accurate calculation is crucial, along with interpreting the results in the context of the study.

Confidence Interval

In regression analysis, after estimating the relationship between variables, it's also important to measure the reliability of these estimates. This is where 'confidence intervals' come into play. A confidence interval gives a range of values for an unknown parameter and is associated with a confidence level, which represents the degree of certainty we have in the interval.

In the provided example, we estimated the average length of time a buyer takes to reach a decision when presented with three alternatives. To express our uncertainty about this estimate, we compute the 95% confidence interval. This interval suggests with 95% confidence that the actual mean decision time falls within the provided range. In educational materials, it is critical to clarify that a wider interval might suggest more uncertainty about the parameter's true value, whereas a narrower interval implies greater precision. Further, students should learn that a confidence interval does not say that the true parameter has a 95% chance of being in the interval; rather, it means that 95% of such intervals will contain the true parameter if we repeated the experiment many times.