Perform the test at the significant level is 0.05. A field biologist examined the sex ratio at birth of the lesser snow geese. A random sample of nests containing four eggs was taken. For each egg resulting in a live gosling, the laying order and the gender of the gosling were recorded. Of the 27 successfully hatched first eggs, 17 were male. Is there significant evidence that the proportion of male goslings is not 50%.

Test if the proportion of male goslings is different of 50%.
At the null hypothesis, it is tested if the proportion is of 50%, that is: [tex]H_0: p = 0.5[/tex]
At the alternative hypothesis, if is tested if the proportion is different of 50%, that is: [tex]H_1: p \neq 0.5[/tex]

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The test statistic is given by:

[tex]z = \frac{X - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which

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The proportion tested is 0.5, thus [tex]p = 0.5[/tex]
The sample proportion is 17 out of 27, thus [tex]n = 27, X = \frac{17}{27} = 0.6296[/tex].

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The value of the test statistic is:

[tex]z = \frac{X - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

[tex]z = \frac{0.6296 - 0.5}{\sqrt{\frac{0.5\times0.5}{27}}}[/tex]

[tex]z = 1.35[/tex]

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Two-tailed test, thus the p-value of the test is P(|z| > 1.35), which is 2 multiplied by the p-value of z = -1.35.
Looking at the z-table, z = -1.35 has a p-value of 0.0885.
The p-value of the test is: [tex]0.0885 \times 2 = 0.177[/tex]

The p-value of the test is 0.177 > 0.05, which means that there is not significant evidence that the proportion of male goslings is not 50%.

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