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bonufazy [111]
3 years ago
13

given that a test statistic in a left tailed test is z equals -1.25, use a 0.05 significance level to find the p-value and state

the conclusion about the null hypothesis

Mathematics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

The <em>p-</em>value of the test is 0.106.

The null hypothesis will be accepted at 5% level of significance.

Step-by-step explanation:

The hypothesis test is left-tailed.

The test statistic value is: <em>z</em> = -1.25.

The significance level of the test is: <em>α</em> = 0.05.

The <em>p</em>-value of a left-tailed hypothesis test is:

p-value=P(Z

The <em>p-</em>value of the test is 0.106.

**Use the <em>z-</em>table for the probability.

<u>Decision rule:</u>

If the <em>p</em>-value is less than the significance level the null hypothesis will be rejected and if it is more than the significance level the null hypothesis will be accepted.

The <em>p</em>-value = 0.106 > <em>α</em> = 0.05.

The null hypothesis will be accepted at 5% level of significance.

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A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
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Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

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3 years ago
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The length of line segment l n on which point m lines in between is 64 units. Option 4 is the correct option.

<h3>What is the length of line segment?</h3>

Length of a line segment is the distance of both the ends of it.

Point m lies between points l and n on line segment l n .

  • The space between l and m is 10x 8.
  • The space between m and n is 5x -4.

The value of line segment LN is,

LN = 12x +16

The sum of LM and LN is equal to the line segment LN. Thus,

LM+MN=LN\\10x +8+(5x-4)=12x+ 16\\10x +8+5x-4=12x+ 16\\10x+5x-12x=16 -8+4\\3x=12\\x=\dfrac{12}{3}\\x=4

Put this value of x in the equation of line segment LN,

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Thus, the length of line segment l n on which point m lines in between is 64 units. Option 4 is the correct option.

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7 0
2 years ago
Students were surveyed about their favorite colors 1/4 preferred red, 1/8 preferred blue, 3/5 preferred green. If 15 students pr
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Answer:

Let us assume the total number of students surveyed = x

Then

Number of students preferring red = x/4

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                                                  = (8x - 3x)/8

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We also know that the total number of students preferring the color green is 15

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3x/8 = 15

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So in total 40 students were surveyed. I hope the procedure is clear to you.

Step-by-step explanation:

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