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snow_tiger [21]
2 years ago
13

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

. A random sample of 300 students taken from this year's student body showed the following number of students in each class.
Freshmen 83
Sophomores 68
Juniors 85
Seniors 64
We are interested in determining whether there has been a significant change in the distribution of class between the last school year and this school year.
A) What is the expected number of freshmen?
B) What is the expected number of seniors?
C) What is the calculated value for the test statistic?
D) What is the p-value?
E) At 95% confidence, should the null hypothesis be rejected?
Mathematics
1 answer:
goblinko [34]2 years ago
8 0

Answer:

1.66

Step-by-step explanation:

We need to conduct a chi square test

The statistic used to check this is given by the below

∑(n, 1) = [(O(i) - E(i)] / E(i)

From the question, we're told the observed value is

Freshmen

Sophomore

Junior

Senior.

After that, we then calculate the expected value.

Freshman

0.3 * 300 = $90

Sophomore

0.24 * 300 = $72

Statistics then finally gives us this

X² = (93 - 90²)/90 + (68 - 72)²/72 + (85 - 78)/78 + (64 -60)²/60

X² = 3² / 90 + 4² / 73 + 13² / 78 + 4² / 60

X² = 0.1 + 0.212 + 2.167 + 0.266 x+

X² = 2.745

X = √2.745

X = 1.66

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Xelga [282]

Answer:

The 98% confidence interval of the proportion = (0.312, 0.374)

Step-by-step explanation:

(Give answers accurate to 3 decimal places.)

The formula for Confidence Interval of Proportion is given as:

p ± z × √p(1 - p)/n

Where p = Proportion = x/n

x = 440

n = 1282

p = 440/1282 = 0.34321372854

Approximately = 0.343

z = z-score of 98 % confidence interval

= 2.326

Confidence Interval =

= 0.343 ± 2.326 × √0.343(1 - 0.343)/1282

= 0.343 ± 2.326 × √0.225351/1282

= 0.343 ± 2.326 × √0.00017578081

= 0.343 ± 2.326 × 0.01325823555

= 0.343 ± 0.03083865589

0.343 - 0.03083865589

= 0.31216134411

Approximately = 0.312

0.343 + 0.03083865589

= 0.37383865589

Approximately to = 0.374

Therefore, the 98% confidence interval of the proportion = (0.312, 0.374)

6 0
3 years ago
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Katena32 [7]

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