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2 years ago

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A college entrance exam company determined that a score of 24 on the mathematics portion of the exam suggests that a student is

ready for college-level mathematics. To achieve this goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 250 students who completed this core set of courses results in a mean math score of 24.5 on the college entrance exam with a standard deviation of 3.3.
a) Do these results suggest that students who complete the core curriculum are ready for college-level mathematics? That is, are they scoring above 24 on the math portion of the exam?
b) State the appropriate null and alternative hypotheses.

1 answer:

seraphim [82]2 years ago

6 0

Answer:

a) The Z -value 2.397 < 2.576 at 99% or 0.01% level of significance

Null hypothesis is accepted at 0.01% level of significance

<em>They score is above 24 on the math portion of the exam</em>

<em>b) </em>

<u><em>Null Hypothesis</em></u>: There is no significance difference between the college level mathematics and math courses in high school

H₀: μ = 24

<u>Alternative Hypothesis: </u>H₁: μ ≠ 24

<u>Step-by-step explanation:</u>

<u><em>Step(i)</em></u>:-

Given random sample 'n' = 250

Given data sample mean x⁻ = 24.5

Standard deviation = 3.3

<u><em>Null Hypothesis</em></u>: There is no significance difference between the college level mathematics and math courses in high school

H₀: μ = 24

<u>Alternative Hypothesis: </u>H₁: μ ≠ 24

test statistic

$Z = \frac{x^{-} - mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{24.5 - 24}{\frac{3.3}{\sqrt{250} } } = \frac{0.5}{0.2087} = 2.397$

a) 99% or 0.01% level of significance

Level of significance ∝ = 0.01

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.01}{2} } = Z_{0.005} =2.576$

The Z -value 2.397 < 2.576 at 99% or 0.01% level of significance

Null hypothesis is accepted at 0.01% level of significance

<em>They score is above 24 on the math portion of the exam</em>

b) 95% or 0.05% level of significance

Level of significance ∝ = 0.05

$Z_{\frac{\alpha }{2} } = Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

The Z -value 2.397 > 1.96 at 95% or 0.05% level of significance

Null hypothesis is Rejected at 0.05% level of significance

<em>They score is below 24 on the math portion of the exam</em>

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