### Understanding the Vertical-Horizontal Illusion#### Explanation and ExperimentAn example of the vertical-horizontal illusion is depicted in the figure above. Although both lines are of equal length, the vertical line appears significantly longer. To explore the intensity of this illusion, a research experiment was conducted. In this experiment, both lines were exactly 10 inches long. Participants saw the image and were informed that the horizontal line was 10 inches long. They were then asked to estimate the length of the vertical line.For this study, a sample of \( n = 25 \) participants was used. The average estimated length of the vertical line was \( M = 12.2 \) inches, with a standard deviation \( s = 1.00 \).#### Figure Description- **Diagram**: The diagram shows two lines of equal length forming a right angle. The vertical line appears longer due to the illusion.#### Tasks and Analysis1. **Hypothesis Testing**: - Conduct a one-tailed hypothesis test with \( \alpha = 0.01 \). - Goal: Prove participants significantly overestimated the true line length. - Note: Accurate estimation would yield a mean \( \mu = 10 \) inches.2. **Effect Size Calculation**: - Compute estimated \( d \) and \( r^2 \), quantifying variance accounted for by the effect.3. **Confidence Interval**: - Construct a 95% confidence interval for the population mean estimated length of the vertical line.This type of illusion and subsequent study not only provides insight into visual perception but also exemplifies statistical methods for hypothesis testing and confidence interval estimation.

The confidence interval are used to measure the interval range within which the true value of the parameter exist. It is calculated using the values sample statistic and margin of error. It is calculate using the formula, x¯±MOE Where MOE is the margin of error, which is the combined value of critical value and the standard error. Given : Average estimation produce mean i.e. μ=10 inches Random sample size, n=25 Average of the sample, x¯=12.2 inches Standard deviation of the sample, s=1 inches a) Claim : The horizontal line is 10 inches long H0 : μ≤10H1 : μ>10 Level of significance is 0.01 Test statistic : t=x¯-μsn=12.2-10125=11 The critical value at n-1=24 degree of freedom at 0.01 level of significance is 2.492 Since the test statistic is greater than critical value, the null hypothesis is rejected. Hence it is concluded that individuals in the sample significantly overestimate the true length of the line.b) Estimate d : Using the calculated test statistic, d=tn=1125=2.2 Thus, the value of d is 2.2 Estimate r2 : Using the calculated values, r2=d2d2+4=2.222.22+4=0.5475 Thus, the effect size is 0.5475 Hence percentage of variance measure the size of effect is 54.75%c) Confidence interval at 95% for the population mean : The t critical value at 95% is 2.064 (From t test table) Now, CI=x¯±t0.05×sn=12.2±2.064×125=12.2±0.4128=12.2-0.4128,12.2+0.4128=11.7872,12.6128 Thus, the confidence interval at 95% for the population mean is 11.7872 and 12.6128