Question

Random samples of 2 brands of almond candy bars were examined to see if one brand contained more almonds than the other. Altogether 18 bars of brand \(\mathrm{A}\) and 22 of brand \(\mathrm{B}\) were tested. The rank total of brand \(A, T_{x}\), turned out to be 274\. Does this indicate a. \(5 \%\) statistically significant difference between the two brands.

Short answer

The Wilcoxon rank-sum test was used to determine if there is a statistically significant difference between the two almond candy bar brands. The calculated test statistic, \(Z = -2.37\), falls outside the range of \(-1.96 < Z < 1.96\). Hence, there is a 5% statistically significant difference between the two brands of almond candy bars.

Step-by-step solution

Calculate the expected value and variance of the rank sum for brand A

First, calculate the expected value and variance of the rank sum for brand A using the following formulas: Expected value, \(E(T_x) = \frac{n_x(n_x + n_y + 1)}{2}\) Variance, \(Var(T_x) = \frac{n_x n_y (n_x + n_y + 1)}{12}\) where \(n_x\) and \(n_y\) are the sample sizes of brands A and B respectively.

Calculate the standardized test statistic

To calculate the standardized test statistic, use the following formula: \(Z = \frac{T_x - E(T_x)}{\sqrt{Var(T_x)}}\)

Compare the calculated test statistic to the critical value

After calculating the test statistic (Z), you will compare it to the critical value from the standard normal table. Since we are interested in a 5% significance level, the critical value for a two-tailed test is: \(Z_{critical} = \pm 1.96\) If the calculated test statistic, \(Z\), falls outside the range of \(-1.96 < Z < 1.96\), it would indicate a 5% statistically significant difference between the two brands.

Determine if the difference is statistically significant

Based on the calculations in previous steps, we can now determine if the difference in the number of almonds in the two brands is statistically significant at a 5% significance level. If the calculated test statistic falls outside the critical value range, then the difference is statistically significant, otherwise, it is not. Now, let's perform the calculations: 1. Calculate the expected value and variance of the rank sum for brand A: \(E(T_x) = \frac{n_x(n_x + n_y + 1)}{2} = \frac{18(18 + 22 + 1)}{2} = \frac{18(41)}{2} = 369\) \(Var(T_x) = \frac{n_x n_y (n_x + n_y + 1)}{12} = \frac{18 \cdot 22 (41)}{12} = 1596\) 2. Calculate the standardized test statistic: \(Z = \frac{T_x - E(T_x)}{\sqrt{Var(T_x)}} = \frac{274 - 369}{\sqrt{1596}} = -2.37\) 3. Compare the calculated test statistic to the critical value: \(-2.37 < -1.96\) Since the calculated test statistic falls outside the range of \(-1.96 < Z < 1.96\), it does indicate a 5% statistically significant difference between the two brands of almond candy bars.

Key concepts

Expected Value

When we talk about the expected value, it’s like discussing the long-run average if we were to repeat an experiment over and over under the same conditions. In our candy bar example, the expected value represents the average rank sum we would expect for brand A's almond content if we could repeat the sampling endlessly.

Mathematically, the expected value for Brand A, commonly denoted as E(T_x), is calculated using the sample sizes of both brands. In the context of our problem, we calculated E(T_x) to be 369, which tells us the average rank sum that would represent no difference between the two brands.

Variance

Now, to grasp the idea of variance: it quantifies the spread or the variability of the ranking sums around the expected value. Essentially, it gives us an insight into how much the rank sums are likely to fluctuate.

In simpler terms, if the variance is high, the rank sums for brand A are not very consistent and vary a lot from the expected value. On the flip side, a low variance indicates that the rank sums are more uniform and cluster closer to the expected value. Our problem presented us with a variance of 1596, which is used to measure the distribution of the data points.

Standardized Test Statistic

Diving into the standardized test statistic, often labeled as Z, transforms our specific sample statistic into a common scale that we can compare against a known distribution – usually a normal distribution. It’s basically telling us how many standard deviations our rank sum is from the expected value.

Calculating a Z-score involves subtracting the expected value from the observed rank sum, then dividing by the standard deviation (which is the square root of the variance). In this case, we got a Z-score of -2.37, indicating that the observed rank sum is about 2.37 standard deviations below what we'd expect if there were no difference in almond content between the two brands.

Critical Value

The critical value is a key concept in hypothesis testing – it’s the threshold against which we compare our standardized test statistic to determine statistical significance.

For a given significance level (in our case, 5%), the critical value marks the point beyond which we start to doubt the null hypothesis (which assumes no difference between groups). With our significance level, we use the critical value of ±1.96. If our Z-score falls outside the ±1.96 range, it suggests the observed differences are too uncommon to be attributed to random chance, pointing towards a significant difference.

Normal Distribution

Normal distribution is a continuous probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In a graph, this distribution will appear as the famous 'bell curve'. The majority of the data falls within a few standard deviations from the mean. This property makes it a fundamental assumption for many statistical procedures including the one in our exercise. Since ranking is often normally distributed, we can use the standard normal table to find critical values and make inferences about our data.