Do birds learn to time their breeding?

- Blue titmice eat caterpillars. The birds would like lots of caterpillars around when they have young to feed, but they breed earlier than peak caterpillar season.
- Do the birds learn from one year's experience when they time breeding the next year?
- Researchers randomly assigned 7 pairs of birds to have the natural caterpillar supply supplemented while feeding their young and another 6 pairs to serve as a control group relying on the natural food supply.
- The next year, they measured how many days after the caterpillar peak the birds produced their nestlings.

The following exercise is based on this experiment:

16. What type of t statistic (paired or two-sample) is this?
- A. Matched pairs statistic.
- B. Two sample statistic.

17. What are the conservative degrees of freedom for this statistic? Give your answer as a whole number.

18. Show that this t leads to the quoted conclusion. Give the P-value for the test.
- A. 0.20 < P < 0.30
- B. 0.10 < P < 0.15
- C. 0.15 < P < 0.20
- D. 0.30 < P < 0.40

19. Does this P-value lead to the quoted conclusion?
- A. Yes
- B. No

Note: In the experimental year, the degree of synchronization did not differ between food-supplemented and control females. For this comparison, the report gives t = -1.05.

A statistic is a numerical summary of a sample, which is a subset of a larger population. Statistics are used in a wide range of fields, including business, economics, social sciences, and more.

What is T statistics?

The t-statistic is a measure of the difference between a sample mean and a population mean, divided by the standard error of the sample mean. It is used in hypothesis testing to determine if the difference between the two means is significant.

According to the given information:

16) This is a two-sample t statistic.

17 )The conservative degrees of freedom for this statistic can be calculated using the formula: df = (n1 - 1) + (n2 - 1), where n1 and n2 are the sample sizes of the two groups. In this case, the sample sizes are 7 and 6, so the degrees of freedom are (7-1) + (6-1) = 11.

18) To show that this t leads to the quoted conclusion, we need to calculate the p-value for the test. Since the t statistic is negative, we will use a one-tailed test to calculate the p-value. Using a t-distribution table or software, we can find that the p-value for a one-tailed t-test with 11 degrees of freedom and a t-statistic of -1.05 is approximately 0.16. Therefore, we can conclude that the difference between the two groups is not statistically significant at the alpha = 0.05 level, and the reported conclusion is correct.

19) Answer: C. 0.15 < P < 0.20

Yes, this P-value leads to the quoted conclusion. The p-value is greater than 0.05, indicating that we fail to reject the null hypothesis of no difference between the two groups, which is consistent with the study report.

20)Question is incomplete

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