Mathematics, 17.07.2021 01:00 confi3353

In a large midwestern university (the class of entering freshmen being on the order of 6000 or more students), an SRS of 100 entering freshmen in 1999 found that 20 finished in the bottom third of their high school class. Admission standards at the university were tightened in 1995. In 2001 an SRS of 100 entering freshmen found that 10 finished in the bottom third of their high school class. The proportion of all entering freshmen in 1999 and 2001, who graduated in the bottom third of their high school class, are p1 and p2, respectively. Is there evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced, as a result of the tougher admission standards adopted in 2000, compared to the proportion in 1999? To determine this, you test the hypothesesH0 : p1 = p2 , Ha : p1 > p2.The P-value of your test isA. 0.976.B. 0.024.C. 0.048.D. 0.001.

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Answer from: michellen2020

B. 0.024

The p-value of the test is 0.024 < 0.05(standard significance level), which means that there is enough evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

Step-by-step explanation:

Before testing the hypothesis, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

1999:

Of 100, 20 were in the bottom thid. So

$p_B = \frac{20}{100} = 0.2$

$s_B = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

2001:

Of 100, 10 were in the bottom third, so:

$p_A = \frac{10}{100} = 0.1$

$s_A = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

To determine this, you test the hypotheses H0 : p1 = p2 , Ha : p1 > p2.

Can also be rewritten as:

H_0: p_B - p_A = 0

H_1: p_B - p_A 0

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the sample:

X = p_B - p_A = 0.2 - 0.1 = 0.1

$s_A = \sqrt{s_A^2+s_B^2} = \sqrt{0.03^2+0.04^2} = 0.05$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{0.1 - 0}{0.05}$

z = 2

P-value of the test and decision:

The p-value of the test is the probability of finding a difference of proportions of at least 0.1, which is 1 subtracted by the p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.976.

1 - 0.976 = 0.024, so the p-value is given by option B.

Answer from: Quest

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.

Answer from: Quest

answer: 2 answers

step-by-step explanation: the first one means to spread the product throughout the marketplace such that a large number of people can buy it.

the second means the process of making a product or service available for the consumer or business user that needs it.

mark this brainliest and i will be glad to .

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