Question

The following are intraocular pressure (mm Hg) values recorded for a sample of 21 elderly subjects: 14.5 13.9 15.0 17.1 12.0 17.5 14.1 12.9 16.9 13.0 17.4 24.2 12.2 14.4 16.0 10.0 18.5 20.8 16.2 15.9 18.6 Can we conclude from these data that the mean of the population from which the sample was drawn is greater than 14? Let α = 0.05. 1. Write the hypotheses, indicate the claim 2. find the critical value t-value 3. calculate the standardized t -value 4. what is the decision

Answer 1

	Values ( X )	$\Sigma (X_{i} - \bar{X})^{2}$
	14.5	1.6045
	13.9	3.4846
	15	0.5878
	17.1	1.7777
	12	14.188
	17.5	3.0043
	14.1	2.7779
	12.9	8.218
	16.9	1.2844
	13	7.6546
	17.4	2.6677
	24.2	71.1205
	12.2	12.7213
	14.4	1.8679
	16	0.0544
	10	33.2548
	18.5	7.4709
	20.8	25.3341
	16.2	0.1877
	15.9	0.0178
	18.6	8.0276
Total	331.1	207.3065

Mean $\bar{X} = \Sigma X_{i} / n$
$\bar{X} = 331.1 / 21 = 15.7667$
Standard deviation $S_{X} = \sqrt{\Sigma (X_{i} - \bar{X})^{2}/n-1}$
$S_{X} = \sqrt{ 207.3065 / 21 -1} = 3.2195$

To Test :-

H0 :-   $\mu = 14$

H1 :-   $\mu > 14$

Test Statistic :-
$t = ( \bar{X} - \mu ) / (S /\sqrt{n})$
$t = ( 15.7667 - 14 ) / ( 3.2195 /\sqrt{ 21 })$
t = 2.5147

Test Criteria :-
Reject null hypothesis if $t \; > \;t_{\alpha, n-1}$
$t_{\alpha, n-1} = t_{0.05 , 21-1} = 1.725$
$t > t_{\alpha, n-1} = 2.5147 > 1.725$
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 2.5147 ) = 0.0103
Reject null hypothesis if P value < $\alpha = 0.05$    level of significance
P - value = 0.0103 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to support the claim that the mean of the population from which the sample was drawn is greater than 14 at α = 0.05.