Respuesta :
Using the z-distribution, the p-value of the test is enough proof to state that freshmen own these devices at a lower rate than juniors.
What are the hypotheses tested?
At the null hypotheses, it is tested if the proportion is of 71%, that is:
[tex]H_0: p = 0.71[/tex]
At the alternative hypotheses, it is tested if the proportion is less than 71%, hence:
[tex]H_1: p < 0.71[/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, we have that the parameters are given as follows:
[tex]\overline{p} = 0.58, p = 0.71, n = 52[/tex].
Hence, the test statistic is given by:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.58 - 0.71}{\sqrt{\frac{0.71(0.29)}{52}}}[/tex]
z = -2.06.
What is the p-value of the test?
Using a z-distribution calculator, with z = -2.06, the p-value(which is the probability of finding a sample proportion of at most as the one found) is of 0.0197.
Since this p-value is less than the standard of 0.05, it is enough proof to state that freshmen own these devices at a lower rate than juniors.
More can be learned about the z-distribution at https://brainly.com/question/16313918
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