6. The t test for two independent samples-One-tailed example using tables Aa Aa
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6. The t test for two independent samples-One-tailed example using tables Aa Aa Most engaged couples expect or at least hope that they will have high levels of marital satisfaction. However because 54% of first marriages end in divorce, social scientists have begun investigating influences on marital satisfaction. [Data source: This data was obtained from National Center for Health Statistics.] Suppose a clinical psychologist sets out to look at the role of economic hardship in relationship longevity. He deci to measure marital satisfaction in a group of couples living e the poverty level and a group of couples living below the poverty level. He chooses the Marital Satisfaction Inventory, because it refers to "partner" and "relationship" rather than "spouse" and "marriage," which makes it useful for research with both traditional and nontraditional couples. Higher scores on the Marital Satisfaction Inventory indicate greater satisfaction. There is c score per couple. Assume that these scores are normally distributed and that the variances of the scores are the same among couples living above the poverty level as among couples living below the poverty level. The psychologist thinks that couples living above the poverty level will have greater relationship satisfaction than couples living below the poverty level. He identifies the null and alternative hypotheses as: Ho·Heouples living above the poverty level Heouples living below the poverty level Hi : couples living above the poverty level->- Heouples living below the poverty level This is a one- tailed test. The psychologist collects the data. A group of 39 couples living above the poverty level scored an average of 51.1 with a sample standard deviation of 9 on the Marital Satisfaction Inventory. A group of 31 couples living below the poverty level scored an average of 45.2 with a sample standard deviation of 12. Use the t distribution table. To u the table, you will first need to calculate the degrees of freedom. The degrees of freedom are 68Explanation / Answer
Given that,
mean(x)=51.1
standard deviation , s.d1=9
number(n1)=39
y(mean)=45.2
standard deviation, s.d2 =12
number(n2)=31
null, Ho: u1 <= u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.668
since our test is right-tailed
reject Ho, if to > 1.668
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (38*81 + 30*144) / (70- 2 )
s^2 = 108.794118
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=51.1-45.2/sqrt((108.794118( 1 /39+ 1/31 ))
to=5.9/2.509797
to=2.350788
| to | =2.350788
critical value
the value of |t | with (n1+n2-2) i.e 68 d.f is 1.668
we got |to| = 2.350788 & | t | = 1.668
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 2.3508 ) = 0.01082
hence value of p0.05 > 0.01082,here we reject Ho
ANSWERS
---------------
null, Ho: u1 <= u2
alternate, H1: u1 > u2
test statistic: 2.350788
critical value: 1.668
decision: reject Ho
p-value: 0.01082
we have enough evidence to support the claim
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