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Sagot :
Incomplete question. However, I answered from a general research perspective.
Explanation:
Note, the term sample used in statistical research refers to subjects or persons carefully selected (using a particular sampling method/type ) to represent the target population.
A randomly selected sample could be done as either:
- Simple Random Sampling
- Stratified Random Sampling
- Cluster Random Sampling.
- Systematic Random Sampling.
Using any of the above sampling methods could result in 70% of the population being from low-income families as this is a matter of chance. Hence, we can agree with the research results.
Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, there is enough evidence to conclude that the children were not randomly selected.
At the null hypothesis, we test if they were really randomly selected, that is, the proportion is of 80%:
[tex]H_0: p = 0.8[/tex]
At the alternative hypothesis, we test if they were not randomly selected, that is, the proportion is of less than 80%.
[tex]H_1: p < 0.8[/tex]
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.
For this problem, the parameters are:
[tex]p = 0.8, n = 150, \overline{p} = 0.7[/tex]
Then, the value of the test statistic is:
[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]z = \frac{0.7 - 0.8}{\sqrt{\frac{0.8(0.2)}{150}}}[/tex]
[tex]z = -3.06[/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, there is enough evidence to conclude that the children were not randomly selected.
A similar problem is given at https://brainly.com/question/24166849
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