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Sagot :
Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, this finding does suggest that the passing rate for teenagers is lower than what CHCCS reported.
Hypothesis
- At the null hypothesis, it is tested if the passing rate is of at least 82%, that is:
[tex]H_0: p \geq 0.82[/tex]
- At the alternative hypothesis, it is tested if the passing rate is of less than 82%, that is:
[tex]H_1: p < 0.82[/tex]
Randomiztion and 10% condition
- The students on the sample are chosen at random, hence the randomization condition is respected.
- 75 students passed, 25 did not, hence there are at least 10 successes and 10 failures in the sample and the 10% condition is also respected.
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 = 0.82, n = 100, \overline{p} = \frac{75}{100} = 0.75[/tex]
Hence, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.75 - 0.82}{\sqrt{\frac{0.82(0.18)}{100}}}[/tex]
[tex]z = -1.82[/tex]
The critical value for a left-tailed test, as we are testing if the proportion is less than a value, with a significance level of 0.05 is of [tex]z^{\ast} = -1.645[/tex]
Since the test statistic is less than the critical value for the left-tailed test, this finding does suggest that the passing rate for teenagers is lower than what CHCCS reported.
To learn more about the z-distribution, you can take a look at https://brainly.com/question/26013190
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