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Sergeu [11.5K]
2 years ago
5

Let μ denote the true average radioactivity level(picocuries per liter). The value 5 pCi/L is considered thedividing line betwee

n safe and unsafe water. Would you recommend testing H0: μ = 5 vs Ha: μ > 5 or H0: μ =5 vs Ha: μ < 5?
Mathematics
1 answer:
VMariaS [17]2 years ago
7 0

Answer:

H0: μ = 5 versus Ha: μ < 5.

Step-by-step explanation:

Given:

μ = true average radioactivity level(picocuries per liter)

5 pCi/L = dividing line between safe and unsafe water

The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.

A type I error, is an error where the null hypothesis, H0 is rejected when it is true.

We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.

Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.

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Lara draws some triangle models with the measurements she wants to use to construct them. Which of these
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Answer: it’s a

Step-by-step explanation:

Cause I think it’s is

6 0
2 years ago
Read 2 more answers
The U.S. Energy Information Administration claimed that U.S. residential customers used an average of 10,608 kilowatt hours (kWh
Marat540 [252]

Answer:

The calculated value Z = 1.4460 < 1.96 at 0.05 level of significance.

The null hypothesis is accepted

A local power company believes that residents in their area use more electricity on average than EIA's reported average

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 10,608 kWh of electricity this year.

Given that the size of the sample n = 187

Given that mean of sample x⁻ = 10737 kWh

The Standard deviation of the Population = 1220kWh

Level of significance = 0.05

The critical value (Z₀.₀₅)= 1.96

<u><em>Step(ii):-</em></u>

Null Hypothesis: H₀:μ > 10608 kWh

Alternative Hypothesis: H₁: μ < 10608kWh

Test statistic

                 Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }

                Z = \frac{10737- 10608 }{\frac{1220}{\sqrt{187} } }

                Z = 1.4460

<u>Final answer:-</u>

The calculated value Z = 1.4460 < 1.96 at 0.05 level of significance.

The null hypothesis is accepted

A local power company believes that residents in their area use more electricity on average than EIA's reported average.

3 0
2 years ago
A student thought she had 3 missing assignments, she actually had 5 missing assignments. What was her percent error?
Licemer1 [7]

Answer:

0.6%

Step-by-step explanation:

8 0
2 years ago
Suppose that a sample of size 100 is to be drawn from a population with standard deviation L0. a. What is the probability that t
NARA [144]

Answer:

The probability that the sample mean will lie within 2 values of μ is 0.9544.

Step-by-step explanation:

Here

  • the sample size is given as 100
  • the standard deviation is 10

The probability that the sample mean lies with 2 of the value of μ is given as

                                            P(| \bar{X}-\mu|

Here converting the values in z form gives

P(-2

Substituting values

P(-2

From z table

P(z\leq 2)=0.9772\\P(z\leq -2)=0.0228\\P(-2\leq z\leq 2)=P(z\leq 2)-P(z\leq -2)\\P(-2\leq z\leq 2)=0.9772-0.0228\\P(-2\leq z\leq 2)=0.9544\\

So the probability that the sample mean  will lie within 2 values of μ is 0.9544.

5 0
3 years ago
Find the unknown side length, x. Write your answer in simplest radical form.
velikii [3]
The answer should be 16√5
8 0
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