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Tamiku [17]
3 years ago
5

A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher a

dministers the test to 200 random students and determines that 151 students pass the exam. The following is the setup for this hypothesis test: {H0:p=0.80 Ha:p>0.80 Find the test statistic for this hypothesis test for a proportion. Round your answer to 2 decimal places.
Provide your answer below: $$

test statistic:
Mathematics
1 answer:
erica [24]3 years ago
6 0

Answer:

z=-1.591

Step-by-step explanation:

Null Hypotheses,

H_0 : p=0.8\\H_a: p>0.80

So we use z-test for one population proportion (right-tailed test)

According to this informaton, we defined that

\alpha=0.05

z_c=1.64 (critical value)

So our Rejection region is R={z:z>1.64}

z=\frac{\bar{p}-p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.755-0.8}{\sqrt{0.8(1-0.8)/200}}=-1.591

Not rejection.

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5 0
3 years ago
34 in.
Troyanec [42]

Answer:

Step-by-step explanation:

Lateral surface area of the triangular prism = Perimeter of the triangular base × Height

By applying Pythagoras theorem in ΔABC,

AC² = AB² + BC²

(34)² = (16)² + BC²

BC = \sqrt{1156-256}

     = \sqrt{900}

     = 30 in.

Perimeter of the triangular base = AB + BC + AC

                                                      = 16 + 30 + 34

                                                      = 80 in

Lateral surface area = 80 × 22

                                  = 1760 in²

Total Surface area = Lateral surface area + 2(Surface area of the triangular base)

Surface area of the triangular base = \frac{1}{2}(\text{Base})(\text{Height})

                                                           = \frac{1}{2}(30)(16)

                                                           = 240 in²

Total surface area = 1760 + 2(240)

                               = 1760 + 480

                               = 2240 in²

Volume = Area of the triangular base × Height

             = 240 × 20

             = 4800 in³

6 0
2 years ago
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