Asked by Paula
it has been found that 85.6% of all enrolled college and university students in the U.S. are undergraduates. A random sample of 500 enrolled college students in a particular state revealed that 420 of them were undergraduates. Is there sufficient evidence to conclude that the population differs from the national percentages? Use a = 0.05
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Writeacher
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Answered by
Brent lincon
It has been found that 85.6% of all enrolled college and university students in tge united states are undergrates. A random sample of 500 enrolled college students in a paticular state revealed that 420 of them were undergrates. Is there sufficient evidence.to conclude that the proporation differs from the national percentage? Use x= 0.05
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Answered by
dyosang gwapa
This is 1-Proportion Z-Test with p = 420 / 500 = 0.84
Ho: mean proportion = 0.856
Ha: mean proportion ≠ 0.856 [Since we are using ≠ , this is a two-tailed test]
z-statistic = (0.84 - 0.856) / sqrt[ (0.856)(1-0.856) / 500] = -1.019
p-value = (2)P(z < -1.019) = (2)(0.1541) = 0.3082
NOTE: we multiplied by 2 because this is a "two-tailed" test.
Conclusion: Since p-value > 0.05, do NOT reject the null hypothesis. There is significant evidence that the mean proportion of undergraduates is not different from 85.6%
hope that helped
Ho: mean proportion = 0.856
Ha: mean proportion ≠ 0.856 [Since we are using ≠ , this is a two-tailed test]
z-statistic = (0.84 - 0.856) / sqrt[ (0.856)(1-0.856) / 500] = -1.019
p-value = (2)P(z < -1.019) = (2)(0.1541) = 0.3082
NOTE: we multiplied by 2 because this is a "two-tailed" test.
Conclusion: Since p-value > 0.05, do NOT reject the null hypothesis. There is significant evidence that the mean proportion of undergraduates is not different from 85.6%
hope that helped
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