Chapter 10: Q.10.127 (page 442)

Glaucoma and Corneal Thickness. Glaucoma is a leading cause of blindness in the United States. N. Ehlers measured the corneal thickness of eight patients who had glaucoma in one eye but not in the other. The results of the study were published as the paper "On Corneal Thickness and Intraocular Pressure, II" (Acta Opthalmologica, Vol. 48, pp. 1107-1112). The data on corneal thickness, in microns, are shown in the following table.
At $10 %$ significance level, do the data provide sufficient evidence to conclude that mean conceal thickness is greater in normal eyes than a eyes with glaucoma.

Short Answer

Expert verified

The null hypothesis is not rejected, and the data are sufficient to indicate that mean corneal thickness in normal eyes is bigger than in glaucoma eyes.

Step by step solution

Given Information

Given data is :

We have to explain whether the data provide sufficient evidence to conclude that mean corneal thickness is greater in normal eyes than in eyes with glaucoma.

Explanation

The null and alternative hypothesis are:

$H_{0} : μ_{1} = μ_{2}$

$H_{a} : μ_{1} > μ_{2}$

The table is shown below:

The mean and standard deviation are computed as follows:

Mean:

$\bar{d} = \frac{\sum d}{n} = \frac{32}{8} = 4$

Standard deviation:

$s_{d} = \sqrt{\frac{\sum d_{i}^{2} - \frac{{(\sum d_{i})}^{2}}{n}}{n - 1}} = \sqrt{\frac{936 - \frac{(32)^{2}}{8}}{8 - 1}} = 10.74$

The formula of test statistics is: $t = \frac{\bar{d}}{\frac{s_{d}}{\sqrt{n}}}$

Substitute the given values:

$t = \frac{4}{\frac{10.74}{\sqrt{6}}} = 1.05$

The degree of freedom is :

$d o f = n - 1 = 8 - 1 = 7$

The critical value for the level of significance is $1.415$

Since, the value of test statistic is not fall in the rejection region.

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