Using the z-distribution, it is found that the test statistic and the p-value of the test are given as follows:
z = –1.78, P-value = 0.0375.
What are the hypothesis tested?
At the null hypothesis, it is tested if the proportion is of 85%, that is:
[tex]H_0: p = 0.85[/tex]
At the alternative hypothesis, it is tested if it is less than that, hence:
[tex]H_1: p < 0.85[/tex].
What are the test statistic and the p-value?
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.
In this problem, we have that the parameters are given as follows:
[tex]n = 50, \overline{p} = 0.76, p = 0.85[/tex]
Hence, the test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.76 - 0.85}{\sqrt{\frac{0.85(0.15)}{50}}}[/tex]
z = -1.78.
Using a z-distribution calculator, considering that we have a left-tailed test, as we are testing if the proportion is less than a value, the p-value is of 0.0375.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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