4. A study of natural selection in a wild bird population compared overwiter sarvival of a subgroup that winters near tree line with the genenl population whose Witer martality is $$ 60 \% $$. Nine of twelve individuals who overwintered nerr ther line navived What is the probability of observing a reult this extrene if ther morality mes sthe sume a ter poel population?

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We assume that the survival probability under the null hypothesis is the same as in the general population. Since the winter mortality is $$60\%$$, the survival probability is

$$
p=1-0.60=0.40.
$$

For a sample of $$n=12$$ birds, the number of survivors $$X$$ follows a binomial distribution:

$$
X \sim \text{Binomial}(n=12,\ p=0.40).
$$

We observed $$9$$ survivors. The probability of obtaining $$9$$ or more survivors is given by

$$
P(X \ge 9)=\sum_{x=9}^{12}\binom{12}{x}(0.40)^x(0.60)^{12-x}.
$$

We now calculate the individual terms:

1. For $$x=9$$:

$$
\binom{12}{9} = 220,\quad (0.40)^9 \approx 0.000262144,\quad (0.60)^3 \approx 0.216.
$$
$$
P(X=9)= 220 \times 0.000262144 \times 0.216 \approx 0.012457.
$$

2. For $$x=10$$:

$$
\binom{12}{10} = 66,\quad (0.40)^{10} \approx 0.0001048576,\quad (0.60)^2 \approx 0.36.
$$
$$
P(X=10)=66 \times 0.0001048576 \times 0.36 \approx 0.002488.
$$

3. For $$x=11$$:

$$
\binom{12}{11} = 12,\quad (0.40)^{11} \approx 0.00004194304,\quad (0.60)^1 = 0.60.
$$
$$
P(X=11)=12 \times 0.00004194304 \times 0.60 \approx 0.000302.
$$

4. For $$x=12$$:

$$
\binom{12}{12} = 1,\quad (0.40)^{12} \approx 0.000016777216,\quad (0.60)^0 = 1.
$$
$$
P(X=12)= 1 \times 0.000016777216 \approx 0.00001678.
$$

Now, summing these probabilities:

$$
P(X \ge 9) \approx 0.012457 + 0.002488 + 0.000302 + 0.00001678 \approx 0.01526378.
$$

Thus, the probability of observing a result this extreme (or more extreme) under the null hypothesis is approximately

$$
\boxed{0.0153}.
$$

This means there is about a $$1.53\%$$ chance of observing $$9$$ or more survivors out of $$12$$ if the true survival probability is $$40\%$$.
The probability of observing 9 or more survivors out of 12 birds, assuming a 40% survival rate, is approximately 1.53%.

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