Does involving a statistician to help with statistical methods improve the chanc
ID: 3231128 • Letter: D
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 (i.e., rejected).
(a) Is this study an experiment or an observational study? Explain.
(b) Arrange these data into a two by two contingency table with the column variable being whether or not statistical help was used.
(c) Compute the percentage of papers not receiving statistical help, that were rejected. Compute the corresponding percentage of papers receiving statistical help, that were rejected. Write a couple of sentences summarizing these results.
(d) Is there a significant difference in the proportions of papers with and without statistical help that are rejected without review? Perform a one-sided test. Write your hypotheses, compute the p-value and state your conclusion.
(e) Does your conclusion in part (a) prove that obtaining statistical assistance increases your chance of getting a paper reviewed? Can you suggest another explanation for your conclusion?
(f) What type of hypothesis testing error might you have made in answering part (c)? Give your answer in terms of medical papers and statistical help.
(g) How often are statisticians involved? Create a 95% confidence interval for the proportion of papers submitted to these journals that include help from a statistician. (Chapter 19)
(h) Create an 80% confidence interval for the difference between the proportion of papers rejected without review when a statistician is and is not involved in the research. Interpret your interval.
Explanation / Answer
(a) Is this study an experiment or an observational study? Explain.
The study is an observation study because there is no experiment involved in this case.
(b) Arrange these data into a two by two contingency table with the column variable being whether or not statistical help was used.
(c) Compute the percentage of papers not receiving statistical help, that were rejected. Compute the corresponding percentage of papers receiving statistical help, that were rejected. Write a couple of sentences summarizing these results.
The papers who doesn't take statistics help are having on average 71% chances of being rejected without review and parallel there are on average 57% chance of being rejected for an paper if it is statistically helped.
(d) Is there a significant difference in the proportions of papers with and without statistical help that are rejected without review? Perform a one-sided test. Write your hypotheses, compute the p-value and state your conclusion.
Null Hypothesis : H0 : There is no difference in rejection without review for papers who take statistically help or not. pstatistichelp = pnothelp
Alternative Hypothesis: Ha : There is significant reduction in rejection rate without review for papers who take statistically help in comparison of who doesn't. pstatistic help < pnot help
Pooled estimate p= (135 + 293)/ (190 + 514) = 0.608
standard error of proportion se0= sqrt [ p* ( 1 - p ) * [ (1/n1) + (1/n2) ] = sqrt [ 0.608 * 0.392 * (1/190 + 1/514)
= 0.04144
Test statistic
Z = (pstatistic help - pnot help)/ se0 = (0.711 - 0.57)/ 0.04144 = 3.40
for p -value = 0.001 so we can reject the null hypothesis and can conclude that there is significant reduction in rejection rate without review for papers who take statistically help in comparison of who doesn't.
e) Does your conclusion in part (d) prove that obtaining statistical assistance increases your chance of getting a paper reviewed? Can you suggest another explanation for your conclusion?
Yes, our conclusion in part(d) prove that obtaining statistical assistance increases your chance of getting a paper reviewed. Here we can also see that Zcritical <Z so we can also reject the null hypothesis.
(f) What type of hypothesis testing error might you have made in answering part (c)? Give your answer in terms of medical papers and statistical help.
There may be a type I error where null hypothesis is true but we have rejected it may be because of bad sampling or non randomness of samples. So it may be true that there is no effect of statistical assistance in lowering in rejection rate of medical papers but we have rejected it.
(g) How often are statisticians involved? Create a 95% confidence interval for the proportion of papers submitted to these journals that include help from a statistician. (Chapter 19)
total papers = 190 + 514 = 704
statisticians involved in 514 paper so proportion = 514/704 = 0.73
95 % confidence interval = p^ +- Z (95%) sqrt [p^ *(1-p^)/N]
= 0.73 +- 1.96 * sqrt [ 0.73 * 0.27/ 704]
= (0.697, 0.763)
so statisticians are normally involved around 70 to 75 % time in submitted papers.
(h) Create an 80% confidence interval for the difference between the proportion of papers rejected without review when a statistician is and is not involved in the research. Interpret your interval.
difference of proportions p1- p2= (0.711 - 0.570) = 0.141
80 % confidence interval = (p1- p2) +- Z (80%) * SE0
= 0.141 +- 1.284 * 0.04144
= (0.088, 0.194)
The interval doesn't consists value of 0 and also have at least 8% difference so it conveys that statistical assistance help papers for not being rejected.
Statistical help used Statistical Help not used Rejected 135 293 Total 190 514Related Questions
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