Question

athletes performing in bright sunlight often smear black eye grease under their eyes to reduce glare. does eye grease work? in one study, 16 student subjects took a test of visual sensitivity to light-and-dark contrast after three hours facing into bright sun, both with and without eye grease. this is a matched pairs design. if eye grease is effective, subjects will be more sensitive to contrast when they use eye grease. the differences in sensitivity, with eye grease minus without eye grease, are provided in the table.
differences in sensitivity
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
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state: we want to know whether eye grease increases sensitivity on the average.
data simulated from a normal distribution based on information in brian m. debroff and patricia j. pahk, \the ability of periorbitally applied antiglare products to improve contrast sensitivity in conditions of sunlight exposure,\ archives of ophthalmology, 121 (2003), pp. 997–1001.
suppose that the subjects are an srs of all young people with normal vision, that contrast differences follow a normal distribution in this population, and that the standard deviation of differences is σ = 0.22. carry out a test of significance.
solve: what is the value of the test statistic? give your answer to two decimal places.
z =
find the p - value of the test. give your answer to four decimal places.
p - value:

Explanation:

Step1: Calculate the sample mean of differences

First, we list the difference values: \(0.07, 0.64, -0.12, -0.05, -0.18, 0.14, -0.16, 0.03, 0.05, 0.02, 0.43, 0.24, -0.11, 0.28, 0.05, 0.29\) (there are 16 values).
The sum of these values:
\(0.07 + 0.64 = 0.71\); \(0.71 - 0.12 = 0.59\); \(0.59 - 0.05 = 0.54\); \(0.54 - 0.18 = 0.36\); \(0.36 + 0.14 = 0.5\); \(0.5 - 0.16 = 0.34\); \(0.34 + 0.03 = 0.37\); \(0.37 + 0.05 = 0.42\); \(0.42 + 0.02 = 0.44\); \(0.44 + 0.43 = 0.87\); \(0.87 + 0.24 = 1.11\); \(1.11 - 0.11 = 1.0\); \(1.0 + 0.28 = 1.28\); \(1.28 + 0.05 = 1.33\); \(1.33 + 0.29 = 1.62\).
The sample mean \(\bar{x}=\frac{1.62}{16}= 0.10125\).

Step2: Calculate the test statistic (z - statistic)

The formula for the z - statistic in a one - sample z - test for a mean (when \(\sigma\) is known) is \(z=\frac{\bar{x}-\mu_0}{\frac{\sigma}{\sqrt{n}}}\).
Here, the null hypothesis \(H_0:\mu = 0\) (eye grease has no effect, so the mean difference is 0), \(\bar{x}=0.10125\), \(\sigma = 0.22\), and \(n = 16\).
\(\frac{\sigma}{\sqrt{n}}=\frac{0.22}{\sqrt{16}}=\frac{0.22}{4}=0.055\)
\(z=\frac{0.10125 - 0}{0.055}=\frac{0.10125}{0.055}\approx1.84\) (rounded to two decimal places)

Step3: Calculate the P - value

Since this is a one - tailed test (we want to know if eye grease increases sensitivity, so we test \(H_1:\mu>0\)), the P - value is \(P(Z > 1.84)\).
From the standard normal table, \(P(Z\leq1.84)=0.9671\), so \(P(Z > 1.84)=1 - 0.9671 = 0.0329\) (rounded to four decimal places)

Answer:

The value of the test statistic \(z\approx\boldsymbol{1.84}\)
The P - value of the test is \(\boldsymbol{0.0329}\)