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2 years ago

Does regents of the university of California v. Bakke remain relevant or has it been superseded by another case?

Advanced Placement (AP)

1 answer:

nikitadnepr [17]2 years ago

7 0

Answer:

It has been suppressed by Grutter v. Bollinger (2003).

Explanation:

According to the University of California v. Bakke case (1978), college applicants’ race was allowed to be a factor in the admission policy, though racial quotas were ruled as impermissible.

Meanwhile, in 2003 Grutter v. Bollinger case ended with a court's decision that admission policy that favors poorly represented ethnic minority groups does not violate the Fourteenth Amendment's Equal Protection Clause, only if the policy takes other factors, such as academic excellence, into account.

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A test of the null hypothesis H0: μ = 0 gives test statistic z = 1.8. a) What is the P-value if the alternative is Ha: μ > μ0

n200080 [17]

According to the result of the test statistic, the p-values are given by:

a) 0.0352.

b) 0.9648.

c) 0.0704.

<h3>How to find the p-value of a test?</h3>

It depends on the test statistic z, as follows.

For a left-tailed test, it is the area under the normal curve to the left of z, which is the p-value of z.

For a right-tailed test, it is the area under the normal curve to the right of z, which is 1 subtracted by the p-value of z.

For a two-tailed test, it is the area under the normal curve to the left of -z combined with the area to the right of z, hence it is 2 multiplied by 1 subtracted by the p-value of z.

In this problem, the test statistic is z = 1.8, which has a p-value of 0.9648.

Item a:

Right-tailed test, hence the p-value is of 1 - 0.9648 = 0.0352.

Item b:

Left-tailed test, hence the p-value is of 0.9648.

Item c:

Two-tailed test, hence the p-value is of 2 x (1 - 0.9648) = 2 x 0.0352 = 0.0704.

You can learn more about p-values at brainly.com/question/13873630

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