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Need help with filling the blanks on this one
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CMV infection and coronary restenosis. Exercise 18.3 found that 43% of CMV+ experienced arterial restenosis within 6 months of atherectomy. In contrast, 8% of CMV- patients experienced a similar outcome. Table 18.8 contains observed counts.
- Repeat the test with a z-statistic. Show the relation between the chi-square statistic and the z-statistic.
[Even though the book asks for the z test, the sample size requirements are not met for the z test. Just proceed anyway, noting the issue]
Hypotheses:
H0: ??
Ha: ??
Assumptions
?????
We use Stata to compute our test statistic and P-value:
. prtesti 49 0.42857 26 0.07692, level(95)
Two-sample test of proportions x: Number of obs = 49
y: Number of obs = 26
------------------------------------------------------------------------------
| Mean Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | .42857 .0706959 .2900085 .5671315
y | .07692 .052258 -.0255038 .1793438
-------------+----------------------------------------------------------------
diff | .35165 .0879137 .1793424 .5239576
| under Ho: .1118792 3.14 0.002
------------------------------------------------------------------------------
diff = prop(x) - prop(y) z = 3.1431
Ho: diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(Z < z) = 0.9992 Pr(|Z| > |z|) = 0.0017 Pr(Z > z) = 0.0008
Results: ???
Interpretation: ??????
Conclusion: ????
Relationship between z stat and chisquare stat: ????
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1. Repeat the test with a z-statistic. Show the relation between the chi-square statistic and the z-statistic.
Hypotheses:
H0: There is no statistically significant difference between the two proportions
Ha: There is a statistically significant difference between the two proportions
Assumptions
i. The information is comprised of simple random values from both populations.
ii. A binomial distribution is followed by both populations.
iii. Samples are independent to one another.
iv. For the test findings to be accurate, np and n(1-p) both should be more than 5.
We use Stata to compute our test statistic and P-value:
. prtesti 49 0.42857 26 0.07692, level(95)
Two-sample test of proportions x: Number of obs = 49
y: Number of obs = 26
------------------------------------------------------------------------------
| Mean Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
x | .42857 .0706959 .2900085 .5671315
y | .07692 .052258 -.0255038 .1793438
-------------+----------------------------------------------------------------
diff | .35165 .0879137 .1793424 .5239576
| under Ho: .1118792 3.14 0.002
------------------------------------------------------------------------------
diff = prop(x) - prop(y) z = 3.1431
Ho: diff = 0
Ha: diff < 0 Ha: diff != 0 Ha: diff > 0
Pr(Z < z) = 0.9992 Pr(|Z| > |z|) = 0.0017 Pr(Z > z) = 0.0008
Results;
Test statistic = 3.1431
P -value = 0.0017
Interpretation;
Since the p value is less than the significance level (0.05), reject the null hypothesis(Ho).
Conclusion;
There is a statistically significant difference between the two proportions
Step-by-step explanation
Relationship between z stat and chi square stat;
Z stat is the same as the chi square test, except we estimate the standard normal deviation instead (z). The test statistic's square (z2) is equivalent to Pearson's chi square statistic(X2).
When the size of the difference between the two proportions is of interest, it is sometimes preferable to the chi square test.
