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Anettt [7]
3 years ago
5

Had to repost it for a better look if still cant see it txt me at 3343729907

Mathematics
1 answer:
kompoz [17]3 years ago
4 0

Answer:

still cant see it sjwjejie

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A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is
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Answer:

The value of the test statistic and its associated p-value at the 5% significance level are -1.54 and 0.9382, respectively.

Step-by-step explanation:

A university interested in tracking its honors program believes that the proportion of graduates with a GPA of 3.00 or below is less than 0.16.

This means that the null hypothesis is:

H_{0}: p \geq 0.16

Testing this hypothesis, means that the alternate hypothesis is:

H_{a}: p > 0.16

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.16 is tested at the null hypothesis:

This means that \mu = 0.16, \sigma = \sqrt{0.16*0.84}

In a sample of 200 graduates, 24 students have a GPA of 3.00 or below.

This means that n = 200, X = \frac{24}{200} = 0.12

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.12 - 0.16}{\frac{\sqrt{0.16*0.84}}{\sqrt{200}}}

z = -1.54

Pvalue:

The pvalue is the probability of finding a sample mean above 0.12, which is 1 subtracted by the pvalue of z = -1.54.

Looking at the z-table, z = -1.54 has a pvalue of 0.0618

1 - 0.0618 = 0.9382

The value of the test statistic and its associated p-value at the 5% significance level are -1.54 and 0.9382, respectively.

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