Answer :
Using the z-distribution, as we are working with a proportion, it is found that there is enough evidence to conclude that the proportion is different of 50%.
What are the hypothesis tested?
At the null hypothesis, it is tested if the proportion is of 0.5, that is:
[tex]H_0: p = 0.5[/tex]
At the alternative hypothesis, it is tested if the proportion is different of 0.5, that is:
[tex]H_1: p \neq 0.5[/tex].
What is the test statistic?
The test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
In which:
- [tex]\overline{p}[/tex] is the sample proportion.
- p is the proportion tested at the null hypothesis.
- n is the sample size.
In this problem, the parameters are:
[tex]p = 0.5, n = 12, \overline{p} = \frac{10}{12} = 0.8333[/tex]
Then, the value of the test statistic is as follows:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.8333 - 0.5}{\sqrt{\frac{0.5(0.5)}{12}}}[/tex]
[tex]z = 2.31[/tex]
What is the decision?
Considering a two-tailed test, as we are testing if the proportion is different of a value, with a significance level of 0.05, the critical value is of [tex]|z^{\ast} = 1.96[/tex].
Since the absolute value of the test statistic is greater than the critical value for the two-tailed test, it is found that there is enough evidence to conclude that the proportion of participants that had improved androstenone-detection accuracy is different of 50%.
More can be learned about the z-distribution at https://brainly.com/question/26454209
