Sandbox: Difference between revisions
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Test of p = 0.5 vs p not = 0.5 | Test of p = 0.5 vs p not = 0.5 | ||
<table width="44%" border="1"> | |||
<tr> | |||
<td width="15%"><div align="center">Sample</div></td> | |||
<td width="6%"><div align="center">X</div></td> | |||
<td width="9%"><div align="center">N</div></td> | |||
<td width="19%"><div align="center">Sample p</div></td> | |||
<td width="25%"><div align="center">95% CI</div></td> | |||
<td width="26%"><div align="center">p-Value</div></td> | |||
</tr> | |||
<tr> | |||
<td><div align="center">1</div></td> | |||
<td><div align="center">69</div></td> | |||
<td><div align="center">165</div></td> | |||
<td><div align="center">0.418182</div></td> | |||
<td><div align="center">0.341979</div></td> | |||
<td><div align="center">0.497378</div></td> | |||
</tr> | |||
</table> |
Revision as of 15:49, 17 December 2007
Of Mice and Males
Authors are not responsible for what journalists write about a research article. Lacking knowledge of statistics, reporters tend to act like stenographers when they aren't extrapolating far beyond the limits of the research. Take a look at what the lay press had to say about "Experimental alteration of litter sex ratios in a mammal" which appeared in the Proceedings of the Royal Society (B).
The Daily Mail:
Red meat and salty snacks are said to lead to boys while chocolate is thought to help to produce girls. Now science suggests the stories may be true: mice with low blood-sugar levels - a good indicator of a sugar-rich diet - produce more female than male offspring.
The Independent:
Boy or girl? Battle of the sexes Are you desperate for a daughter or dying for a son? The solution could lie in a mother's diet - before she even conceives.
New Scientist:
Findings lend credence to traditional beliefs that eating certain foods can influence the sex of offspring.
Discover:
The Biology of . . . Sex Ratios. Want a boy at all costs? The secret may lie in your glucose levels.
FoxNews.com:
Can what a mother-to-be eats influence the sex of her unborn baby? Maybe, says new research.
The research itself looks at a very important issue in biology: the influence of nutrition on reproductive strategy and the ensuing evolutionary advantage. To carry out their research, they had 20 female mice in a control group and 20 female mice in the treatment group which was given "a steroid [DEX] that inhibits glucose transport and reduces plasma glucose concentrations." The original paper does not give a table whereby for each of the 40 mice is recorded the number in the litter, number of males and which arm of the study it was in. Instead, we have to relay on the given summary data: average litter size for control is 10.45 with a standard error of .60, and the average litter size for the treatment is 9.17 with a standard error of .62.
According to the article, "The sex ratio differed significantly between the treatment and control groups (rank-sum test: Z= -2.18, p=0.03), with DEX females giving birth to fewer sons (41.9%) than control females (53.5%)." With this information, it would appear that the control group produced a total of 10.45 * 20 = 209 mice resulting in 209*.535 = 112 males. The treatment group is more difficult to determine because two of the 18 "failed to conceive;" thus, if only 18 are relevant, then the treatment group has 9.17 * 18 = 165 mice and 165 * .419 = 69 males. Using these numbers, a Minitab printout yields a (Fisher exact because of the relatively small samples) p-value of .029 which is close to the "p=.03" mentioned in the article.
Test and CI for Two Proportions
Sample |
X |
N |
Sample p |
1 |
112 |
209 |
.0.535885 |
2 |
69 |
165 |
0.418182 |
Difference = p (1) - p (2)
Estimate for difference: 0.117703
95% CI for difference: (0.0165306, 0.218876)
Test for difference = 0 (vs not = 0): Z = 2.28 P-Value = 0.023
Fisher's exact test: P-Value = 0.029
Discussion
1. No confidence interval for the difference in proportion of males is given in the article itself. Does the 95% CI suggest any guarantee for reduction in male mice? Male humans?
2. Regarding the treatment arm, the article states : "42%, two-tailed binomial test, p=.04." Using the summary data, Minitab reports
Test and CI for One Proportion
Test of p = 0.5 vs p not = 0.5
Sample |
X |
N |
Sample p |
95% CI |
p-Value |
1 |
69 |
165 |
0.418182 |
0.341979 |
0.497378 |