Step 1 :The problem is asking for a 95% confidence interval for the difference in sensitivity to contrast with and without eye grease. This is a problem of estimating the mean of a population based on a sample. The sample in this case is the differences in sensitivity to contrast with and without eye grease.
Step 2 :The confidence interval for the mean of a population is given by the formula: \(\bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}\) where: \(\bar{x}\) is the sample mean, \(t_{\alpha/2, n-1}\) is the t-score for a two-tailed test with \(\alpha\) level of significance and \(n-1\) degrees of freedom, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 3 :In this case, \(\alpha = 0.05\) (since we want a 95% confidence interval), and the sample size \(n = 16\).
Step 4 :First, we need to calculate the sample mean and the sample standard deviation. Then, we can use these values to calculate the confidence interval.
Step 5 :The differences in sensitivity to contrast with and without eye grease are [0.07, 0.64, -0.12, -0.05, -0.18, 0.14, -0.16, 0.03, 0.05, 0.02, 0.43, 0.24, -0.11, 0.28, 0.05, 0.29].
Step 6 :The sample mean is 0.10125 and the sample standard deviation is 0.2263294059551255.
Step 7 :The t-score for a two-tailed test with 0.05 level of significance and 15 degrees of freedom is 2.131449545559323.
Step 8 :Substituting these values into the formula, we get the lower bound of the confidence interval as -0.01935242736744093 and the upper bound as 0.22185242736744093.
Step 9 :Final Answer: The 95% confidence interval for the difference in sensitivity to contrast with and without eye grease is \(\boxed{[-0.0194, 0.2219]}\) to four decimal places. This means we are 95% confident that the true mean difference in sensitivity to contrast with and without eye grease is between -0.0194 and 0.2219.