In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 32 of the 44 online courses at this particular community college were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% repre
1 answer
Question:
In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 32 of the 44 online courses at this particular community college were taught by full-time faculty.
Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question.
Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question.
Answers
Answer:The p-value of the test is 0.5028>0.05, which means that there is sufficient evidence to determine that 68% represents the state in question.Step-by-step explanation:Test if 68% represents the state in question.This means that at the null hypothesis, we test if the proportion is 68%, that is:[tex]H_0: p = 0.68[/tex]At the alternate hypothesis, we test if the proportion is different of 68%, that is:[tex]H_a: p \neq 0.68[/tex]The test statistic is:[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.0.68 is tested at the null hypothesis:This means that [tex]\mu = 0.68, \sigma = \sqrt{0.68*0.32}[/tex]32 of the 44 online courses at this particular community college were taught by full-time faculty.This means that [tex]n = 44, X = \frac{32}{44} = 0.7273[/tex]Test statistic:[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex][tex]z = \frac{0.7273 - 0.68}{\frac{\sqrt{0.68*0.32}}{\sqrt{44}}}[/tex][tex]z = 0.67[/tex]P-value of the test and decision:The p-value of the test is the probability of finding a sample proportion differing by the population proportion by at least 0.7273 - 0.68 = 0.0473, which is P(|z| > 0.67), which is 2 multiplied by the p-value of z = -0.67.Looking at the z-table, z = -0.67 has a p-value of 0.25142*0.2514 = 0.5028The p-value of the test is 0.5028>0.05, which means that there is sufficient evidence to determine that 68% represents the state in question.
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