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A student believes that less than 50% of students at his college receive financial aid. A random sample of 120 students was taken. Sixty-five percent of the students in the sample receive financial aid. Test the hypothesis at the 2% level of significance. What are the p-value and conclusion?

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Answer:

p= 0.9995 and we can conclude that students receiving aid is more than 50% according to the sample results in 2% significance level.

Step-by-step explanation:

[tex]H_{0}[/tex]: less than or equal 50% of students at his college receive financial aid

[tex]H_{a}[/tex]: more than 50% of students receive financial aid.

According to the nul hypothesis we assume a normal distribution with proportion 50%.

z-score of sample proportion can be calculated using the formula:

z=[tex]\frac{X-M}{\frac{s}{\sqrt{N} } }[/tex] where

  • X is the sample proportion (0.65)
  • M is the null hypothesis proportion (0.5)
  • s is the standard deviation of the sample ([tex]\sqrt{0.5*(1-0.5)}[/tex])
  • N is the sample size (120)

then z=[tex]\frac{0.65-0.50}{\frac{\sqrt{0.25}}{\sqrt{120} } }[/tex] ≈ 3,29

thus p= 0.9995 and since p<0.02 (2%), we reject the null hypothesis.

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