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miv72 [106K]
2 years ago
9

Is there any systematic tendency for part-time college faculty to hold their students to different standards than do full-time f

aculty? An article reported that for a sample of 125 courses taught by full-time faculty, the mean course GPA was 2.7386 and the standard deviation was 0.65342, whereas for a sample of 88 courses taught by part-timers, the mean and standard deviation were 2.8439 and 0.49241, respectively. Does it appear that true average course GPA for part-time faculty differs from that for faculty teaching full-time? Test the appropriate hypotheses at significance level 0.01. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
Mathematics
1 answer:
alexandr1967 [171]2 years ago
6 0

Answer:

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

-1. 34

0.1837

Step-by-step explanation:

Full time :

n1 = 125

x1 = 2.7386

s1 = 0.65342

Part time :

n2 = 88

x2 = 2.8439

s2 = 0.49241

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

Test statistic :

The test statistic :

(x1 - x2) / sqrt[(s1²/n1 + s2²/n2)]

(2.7386 - 2.8439) / sqrt[(0.65342²/125 + 0.49241²/88)]

−0.1053 / sqrt(0.0034156615712 + 0.0027553)

-0.1053 /0.0785554

= - 1.34

Test statistic = - 1.34

The Pvalue :

Using df = smaller n - 1 = 88 - 1 = 87

Pvalue from test statistic score ;

Pvalue = 0.1837

Pvalue > α ; We fail to reject the null and conclude that the GPA does not differ.

At α = 0.01 ; the result is insignificant

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2 years ago
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Substitute 2m + n =0 and m + 2n =3
melamori03 [73]

There are several ways you can solve this problem if you're trying to solve for m and n. You can substitute, or systems of equations. However, I'm going to use substitution:

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We can input that in for the other equation:

m + 2n = 3 now becomes: m + 2(-2m) = 3

Now we can solve:

m + 2(−2m) = 3

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However, if you were solving for m+n:

You would add the two equations!:

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