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Vikentia [17]
3 years ago
9

Suppose the mean GPA of all students graduating from a particular university in 1975 was 2.40. The registrar plans to look at re

cords of students graduating last year to see if the mean GPA has changed. Let u = the true mean GPA.
Choose the correct the null and alternative hypotheses for this investigation.

A) H0:mu != 2.30 ; H0 mu = 2.30
B) H0:mu = 2.30 ; H0 mu! = 2.30
C) H0:mu = 2.30 ; H0 mu > 2.30
D) H0:mu = 2.30 ; H0 mu < 2.30
E) H0:mu > 2.30 ; H0 mu = 2.30
F) H0:mu < 2.30 ; H0 mu = 2.30
Mathematics
1 answer:
ycow [4]3 years ago
6 0

Answer:

Correct option:

(B) <em>H₀</em>: <em>μ</em> = 2.40 vs. <em>Hₐ</em>: <em>μ</em> ≠ 2.40.

Step-by-step explanation:

The registrar of particular university in 1975 plans to look at records of students graduating last year to see if the mean GPA has changed from 2.40.

The registrar can use a single mean test to determine whether the mean has changed or not.

The hypothesis can be described as:

<em>H₀</em>: The mean GPA is 2.40, i.e. <em>μ</em> = 2.40.

<em>Hₐ</em>: The mean GPA is different from 2.40, i.e. <em>μ</em> ≠ 2.40.

To perform the test the registrar can either use a <em>z</em>-distribution or a <em>t</em>-distribution.

If the data provided gives some insight about the population standard deviation and the sample selected is quite large then the <em>z</em>-distribution can be used.

Otherwise it is wiser to use a <em>t</em>-distribution.

The decision rule is:

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis will be rejected and vice-versa.

Thus, the correct option is (B).

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Aneli [31]

Using the normal distribution, it is found that 0.2119 = 21.19% of rainy days have rainfall with pH below 5.0.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
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The mean and the standard deviation are given, respectively, by:

\mu = 5.43, \sigma = 0.54

The proportion of rainy days have rainfall with pH below 5.0 is the <u>p-value of Z when X = 5</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{5 - 5.43}{0.54}

Z = -0.8

Z = -0.8 has a p-value of 0.2119.

0.2119 = 21.19% of rainy days have rainfall with pH below 5.0.

More can be learned about the normal distribution at brainly.com/question/25800303

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