Question

Does involving a statistician to help with statistical methods improve the chance that a medical research paper will be published? A study of papers submitted to two medical journals found that 135 of 190 papers that lacked statistical assistance were rejected without even being reviewed in detail. In contrast, 293 of the 514 papers with statistical help were sent back without review. Find the value of the z-test statistic. Give your answer to two decimal places.

Answer 1

z-test statistic is 3.38

Explanation:

Null and alternative hypotheses are:

`H_(0)`: the proportion of papers that lacked statistical assistance sent back without review is the same as the proportion of papers with statistical help sent back without review

`H_(a)`: the proportion of papers that lacked statistical assistance sent back without review is different than the proportion of papers with statistical help sent back without review

Test statistic can be found using the equation:

`z=\frac{p1-p2}{\sqrt{{p*(1-p)*((1)/(n1) +(1)/(n2)) }}}` where

• p1 is the sample proportion of papers that lacked statistical assistance sent back without review ( (
  `(135)/(190)=0.71`)

• p2 is the sample proportion of papers with statistical help sent back without review (
  `(293)/(514)=0.57`)

• p is the pool proportion of p1 and p2 (
  `(135+293)/(190+514)=0.61`)

• n1 is the sample size of papers that lacked statistical assistance (190)

• n2 is the sample size of papers with statistical help (514)

Then
`z=\frac{0.71-0.57}{\sqrt{{0.61*0.39*((1)/(190) +(1)/(514)) }}}` ≈ 3.38