Question

8. In 1979, 15% of primary liver cancer patients had cancer that would go into remis- sion. A new treatment using radioactive antibodies had 42 of 84 patients go into remission. Using the above output, test whether the proportion of remissions for the new treatment is higher than 0.15.

Answer 1

Given : n=84 , X=42

The estimate of the sample proportion is , $\hat{p}=\frac{X}{n}=\frac{42}{84}=0.5$

The null and alternative hypothesis is ,

$H_0:p=0.15$

$H_a:p>0.15$ (Claim)

The test is right-tailed test.

The critical value is , $Z_{\alpha}=Z_{0.05}=1.64$ ; From Z-table

The decision rule is ,

Reject Ho , if Z-stat>1.64 , otherwise fail to reject Ho.

The test statistic is ,

$Z_{stat}=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\frac{0.5-0.15}{\sqrt{\frac{0.15(1-0.15)}{84}}}=8.98$

Decision : Here , Z-stat=8.98>1.64

Therefore , reject Ho.

Conclusion : There is sufficinet evidence to conclude that the proportion of remission for the new treatment is higher than 15%.