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Rina8888 [55]
3 years ago
5

Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors

. A sample of 300 students taken from this year's student body showed the following number of students in each classification.
Freshmen 83
Sophomores 68
Juniors 85
Seniors 64
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year.

Mathematics
1 answer:
Natalija [7]3 years ago
6 0

Answer:

There has been no change in the classifications between the last school year and this school year.

Step-by-step explanation:

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H</em>₀: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

\chi^{2}=\sum\limits^{n}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}

The expected values are computed using the formula:

E_{i}=p_{i}\times N

Here, <em>N</em> = 300

Use Excel to compute the values.

The test statistic value is:

\chi^{2}=\sum\limits^{n}_{i=1}\frac{(O_{i}-E_{i})^{2}}{E_{i}}=1.662

The test statistic value is, 1.662.

The degrees of freedom of the test is:

<em>n</em> - 1 = 4 - 1 = 3

The significance level is, <em>α</em> = 0.05.

Compute the p-value of the test as follows:

<em>p</em>-value = 0.6454

*Use a Chi-square table.

p-value = 0.6454 > <em>α</em> = 0.05.

So, the null hypothesis will not be rejected at 5% significance level.

Thus, concluding that there has been no change in the classifications between the last school year and this school year.

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Read 2 more answers
8.6.PS-14
s2008m [1.1K]

Answer:

Answer:

11 feet

Step-by-step explanation:

Area formed by the watery pattern is 379.94 square feet.

Since the water is being sent out in circular pattern, the maximum distance at which the water is reaching out will be equal to the radius of the circle as the sprinkler is at the center of the circle.

Area of a circle is given as:

\begin{gathered}Area=\pi r^{2}\\\\ 379.94 = \pi r^{2}\\\\r^{2}=\frac{379.94}{\pi}\\\\ \text{Using pi equal to 3.14, we get}\\\\ r^{2}=121\\\\ r = 11\end{gathered}

Area=πr

2

379.94=πr

2

r

2

=

π

379.94

Using pi equal to 3.14, we get

r

2

=121

r=11

This means the radius of the circular pattern being formed is 11 feet. So the sprinkler can spread the water 11 feet away from it.

Step-by-step explanation:

im not sure if that is correct

comment if wrong

*hope it help:))

3 0
3 years ago
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