Question

Suppose that a case-control study was conducted among men in the United States in order to find out whether a mother’s use of hormones during pregnancy influenced her son’s risk of developing testicular cancer later in life. Investigators selected 500 cases who were hospitalized for testicular cancer and 1000 controls. The study found that 90 cases’ mothers and 50 controls’ mothers had used hormones during pregnancy.

Answer 1

Step-by-step explanation:

Using the information given, we can construct a 2x2 contingency table to summarize the data:

```

| Cases | Controls |

---------------|----------|----------|

Hormone use | 90 | 50 |

No hormone use | 410 | 950 |

---------------|----------|----------|

Total | 500 | 1000 |

```

From this contingency table, we can calculate the odds ratio (OR) and the 95% confidence interval (CI) for the association between a mother's use of hormones during pregnancy and her son's risk of developing testicular cancer later in life.

The odds ratio is calculated as the ratio of the odds of exposure (hormone use) in cases to the odds of exposure in controls:

OR = (90/410) / (50/950) = 2.78

This indicates that the odds of hormone use during pregnancy are 2.78 times higher among mothers of sons with testicular cancer compared to mothers of healthy sons.

To calculate the 95% confidence interval (CI) for the odds ratio, we can use the following formula:

ln(OR) ± 1.96 × SE(ln(OR))

where SE(ln(OR)) is the standard error of the natural logarithm of the odds ratio, which can be calculated as:

SE(ln(OR)) = sqrt(1/90 + 1/50 + 1/410 + 1/950) = 0.321

Plugging in the values, we get:

ln(2.78) ± 1.96 × 0.321

= 1.027 ± 0.63

= (0.40, 1.65)

Therefore, the 95% confidence interval for the odds ratio is 0.40 to 1.65. Since this interval includes 1, we cannot conclude with 95% confidence that there is a significant association between a mother's use of hormones during pregnancy and her son's risk of developing testicular cancer later in life.