Answer :
Final answer:
The conditions for inference are met for conducting a z-test for one proportion.Therefore yes, the random, 10%, and large counts conditions are all met is correct.
Explanation:
For conducting a z-test for one proportion, the conditions for inference that need to be met are:
The random condition: The sample should be a random sample from the population of interest. In this case, a random sample of 80 high school students is selected, so the random condition is met.
The 10% condition: The sample size should be less than 10% of the population. In this case, the sample size is 80, which is less than 10% of the population of high school students. Therefore, the 10% condition is met.
The large counts condition: Both the number of successes (students working part-time jobs) and failures (students not working part-time jobs) should be at least 10. In this case, the sample includes 30 students who work part-time jobs (37.5% of 80 students), which is greater than 10. Therefore, the large counts condition is met.
Based on these conditions, we can conclude that the conditions for inference are met for conducting a z-test for one proportion.
The required, there is convincing evidence that the proportion of all high school students who work a part-time job during the school year is greater than 0.28.
The conditions for inference for conducting a z-test for one proportion are:
Random: The sample is selected using a random method, so this condition is met.
- 10%: The sample size (80) is less than 10% of the total population of high school students, so this condition is met.
- Large Counts: Both np and n(1-p) are greater than or equal to 10, where n is the sample size and p is the hypothesized proportion. In this case, np = 80 × 0.28 = 22.4 and n(1-p) = 80 × (1 - 0.28) = 57.6. Since both values are greater than 10, this condition is also met.
Therefore, all the conditions for inference are met, and we can conduct a z-test for one proportion to test whether the proportion of all high school students who work a part-time job during the school year is greater than 0.28.
The null hypothesis is that the true proportion is 0.28, and the alternative hypothesis is that the true proportion is greater than 0.28. We can calculate the test statistic using the formula:
z = (p - P) / √[P(1-P) / n]
where p is the sample proportion (0.375), P is the hypothesized proportion (0.28), and n is the sample size (80).
Plugging in the values, we get:
z = (0.375 - 0.28) / √[0.28 × (1 - 0.28) / 80] = 2.22
Using a standard normal distribution table or calculator, we find that the p-value for a z-score of 2.22 is approximately 0.014. Since this is less than the significance level of 0.05, we reject the null hypothesis and conclude that there is convincing evidence that the proportion of all high school students who work a part-time job during the school year is greater than 0.28.
Learn more about hypothesis tests here:
https://brainly.com/question/30588452
#SPJ1
