Problem 7 - An Introduction to Psychological Statistics

Question

You and your friend are debating which type of candy is the best. You find data on the average rating for hard candy (e.g. jolly ranchers, \(\overline{\mathrm{X}}=3.60\) ), chewable candy (e.g. starburst, \(\overline{\mathrm{X}}=4.20\) ), and chocolate (e.g. snickers, \(\overline{\mathrm{X}}=4.40\) ); each type of candy was rated by 30 people. Test for differences in average candy rating using \(\mathrm{SSB}=16.18\) and \(\mathrm{SSW}=28.74\).

Step-by-Step Solution

Answer

Significant differences exist in the average ratings of the candy types.

Step 1: Understand the Objective

We're testing for differences in average candy ratings across three types: hard candy, chewable candy, and chocolate. Given are sample means and sample sizes, along with the sum of squares between (SSB) and sum of squares within (SSW). We'll perform a one-way ANOVA test to see if any differences in ratings are statistically significant.

Step 2: State the Hypotheses

In a one-way ANOVA, the null hypothesis (H_0) states that the means of all groups are equal, i.e., \( H_0: \mu_{1} = \mu_{2} = \mu_{3} \).The alternative hypothesis (H_a) is that at least one group mean is different.

Step 3: Calculate the F-ratio

The formula for the F-ratio in a one-way ANOVA is:\[ F = \frac{SSB/(k-1)}{SSW/(N-k)} \]where - *SSB* is the Sum of Squares Between,- *SSW* is the Sum of Squares Within,- k is the number of groups, and- N is the total number of observations.Plugging in the provided values: \[ k = 3,\quad N = 90 \,(3 \times 30) \]Calculate:\[F = \frac{16.18/2}{28.74/87} = \frac{8.09}{0.330} \approx 24.52\]

Step 4: Determine the Critical F-value

For k=3 and N-k = 87, use an F-distribution table to find the critical F-value at a chosen significance level, commonly \( \alpha = 0.05 \). For (df1 = 2, df2 = 87) , the critical F-value is approximately 3.10.

Step 5: Compare the F-ratio with Critical F-value

If the calculated F-ratio (24.52) is greater than the critical F-value (3.10), we reject the null hypothesis. In this case, because 24.52 > 3.10, we reject H_0 .

Step 6: State the Conclusion

Since we rejected the null hypothesis, this means there are statistically significant differences in the average ratings of at least one type of candy. We do not yet know which specific candy types significantly differ from each other, but we know not all means are equal.

Key Concepts

Statistical Hypothesis Testing

Statistical hypothesis testing is a process to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. In this exercise, our goal is to explore if there are significant differences in the average ratings of three types of candy: hard candy, chewable candy, and chocolate. We use a one-way ANOVA test to analyze this.
A hypothesis test includes two complementary statements:

• The null hypothesis ( H0): This is the default assumption. For this test, it states that all group means are equal, meaning there's no difference in candy preference.
• The alternative hypothesis ( Ha): This suggests that at least one group mean is different, indicating variability in candy ratings.

The one-way ANOVA helps us determine which hypothesis the data supports by assessing the variance between and within groups.

F-ratio Calculation

The F-ratio is the statistic used in ANOVA tests to compare variances and determine significant differences among group means. It is essential to understand what these variances mean:
The calculation for the F-ratio is:\[ F = \frac{SSB/(k-1)}{SSW/(N-k)} \]

• *SSB* (Sum of Squares Between): The variance among the group means.
• *SSW* (Sum of Squares Within): The variance within each group.
• *k*: The number of groups, which in this case is 3.
• *N*: The total number of observations, being 90 (30 ratings per candy).

Calculating the F-ratio reveals whether the variance between the candy types is significantly greater than the variance within each candy type. For this exercise, the calculated F-ratio is approximately 24.52, which indicates substantial differences if it surpasses the critical F-value.

Critical F-value

The critical F-value is a threshold determined by the degrees of freedom, which helps in deciding whether to reject the null hypothesis or not. First, you need to select a significance level ( \( \alpha \)), commonly 0.05, to determine how much variance you are willing to accept as random.
The degrees of freedom needed are:

• *df1*: Number of groups minus one, (k - 1)
• *df2*: Total observations minus the number of groups, (N - k)

For this problem, df1 = 2 and df2 = 87. Using an F-distribution table or calculator, the critical F-value for these degrees at \( \alpha = 0.05 \) is approximately 3.10. If our calculated F-ratio is larger than this value, it suggests significant differences, leading to the rejection of the null hypothesis.

Sum of Squares

Sum of squares is a crucial concept in ANOVA, used to quantify the variance in data.
Two main types of sum of squares are relevant:

• **Sum of Squares Between (SSB):** Measures how much group means differ from the overall mean. Larger values indicate greater group differences.
• **Sum of Squares Within (SSW):** Reflects variance within a group. This measures how individual observations within the same group diverge from that group's mean.

In our exercise, SSB is 16.18, and SSW is 28.74. These values are crucial in deriving the F-ratio in a way that balances the extent of variance between groups relative to the variance within groups. Proper understanding of these will help you gauge the influence and comparison of variability underpinning the differences in candy ratings.