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omeli [17]
2 years ago
13

Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the 0.10 significance level. T

he null and alternative hypothesis would be: H 0 : p N ≥ p D H 1 : p N < p D H 0 : p N ≤ p D H 1 : p N > p D H 0 : p N = p D H 1 : p N ≠ p D H 0 : μ N ≤ μ D H 1 : μ N > μ D H 0 : μ N ≥ μ D H 1 : μ N < μ D H 0 : μ N = μ D H 1 : μ N ≠ μ D The test is: two-tailed right-tailed left-tailed The sample consisted of 30 night students, with a sample mean GPA of 3.34 and a standard deviation of 0.02, and 30 day students, with a sample mean GPA of 3.32 and a standard deviation of 0.08. The test statistic is: (to 2 decimals) Use the conservative degree of freedoms. The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesis
Mathematics
1 answer:
o-na [289]2 years ago
5 0

Answer:

H0 : μN ≤ μD

H1 : μN > μD

Right tailed

Test statistic = 1.33

Pvalue = 0.097

Fail to reject the Null

Step-by-step explanation:

H0 : μN ≤ μD

H1 : μN > μD

The test is right tailed ; culled from the direction of the greater than sign ">"

Night students :

n1 =30 x1= 3.34 s1 = 0.02

Day students:

n2 = 30 x2 = 3.32 s2 = 0.08

The test statistic :

(x1 - x2) / √(s1²/n1) + (s2²/n2)

T= (3.34 - 3.32) / √(0.02²/30) + (0.08²/30)

T = 0.02 / 0.0150554

Test statistic = 1.328

Using the conservative approach ;

df = Smaller of n1 - 1 or n2 - 1

df = 30 - 1 = 29

Pvalue(1.328, 29) = 0.097

At α = 0.10

Pvalue < α ; Hence, we reject H0 ; and conclude that there is significant evidence that GPA of night student is greater than GPA of day student

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3 years ago
Find the formula for an exponential function that passes through the two points give.
Ksenya-84 [330]

Answer:

y=3(2)^{x}

Step-by-step explanation:

Given.

Two points are given.

(x, y)=(-1,\frac{3}{2}) and (x, y)=(3,24)

An exponential function is in the general form.

y=a(b)^{x}-------(1)

We know the points  (-1,\frac{3}{2}) and (3,24)

put the first point value in equation 1

\frac{3}{2}=a(b)^{-1}

\frac{3}{2}=\frac{a}{b}

a=\frac{3}{2}\times b--------(2)

put the second point value in equation 1

24=a(b)^{3}----------(3)

Put the a value from equation 2 to equation 3

24=\frac{3}{2}\times b(b)^{3}

b^{3+1}=\frac{24\times 2}{3}

b^{4} = 16\\b=\sqrt[4]{16} \\b=2

Put the b value in equation 2

a=\frac{3}{2}\times 2

a=3

Put the a and b value in equation 1

y=3(2)^{x}

So, the exponential function that passes through the points  (-1,\frac{3}{2}) and (3,24)) are y=3(2)^{x}.

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Lexiva is available in 350 mg tablets. How many mg will the patient take in 22 days
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Answer:

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Step-by-step explanation:

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iragen [17]
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The graph of y=cos⁡x is transformed to y=a cos ⁡(x−c)+d by a vertical compression by a factor of 1/3 and a translation 2 units d
rodikova [14]

ANSWER

y =  \frac{1}{3}  \cos(x)  - 2

EXPLANATION

If the graph of y=cos⁡x is transformed to y=a cos ⁡(x−c)+d by a vertical compression by a factor of 1/3 and a translation 2 units down,

then

a=1/3

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Therefore the new equation is:

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Read 2 more answers
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