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Sagot :
Using an hypothesis test, it is found that:
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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- 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
- p is the proportion tested.
- X is the sample proportion.
- n is the size of the sample.
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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%.
A similar problem is given that https://brainly.com/question/24166849
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