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Nady [450]
3 years ago
12

= - 11 \\ " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
irakobra [83]3 years ago
7 0

Answer:

n is equal to -8

Step-by-step explanation:

In order to solve this, we simply need to add 3 to both sides of the equation.

n - 3 = -11

n - 3 + 3 = -11 + 3

n = -8

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Answer: Frequencies .

Step-by-step explanation:

In statistics theory the chi-square test statistic is calculated by considering each cell and the expected frequency is subtracted from the observed frequency after that the difference is squared, and the total is divided by the expected frequency. This is highlight that the key focus of the chi-square test is the frequencies .

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3 years ago
In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were re
11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

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Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

df = 230

We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

Therefore, at a 5% significant level, this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed

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