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juin [17]
2 years ago
12

Assume that a random sample of 4 Goldfinch birds are going to be used to test the above hypothesis. Compute the largest value of

the sample mean, that will allow the experimenter reject the null and prove the alternative at significance level 0.01. For this problem, make sure that your solution includes all steps of your computation. You will not earn points if you just use a formula
Mathematics
1 answer:
Misha Larkins [42]2 years ago
4 0

This question is incomplete, the complete question is;

Assume that body masses of Goldfinch birds follow a normal distribution, with a standard deviation equal to 0.04 oz. An ornithologist who would like to make some inference about the average body mass of the Goldfinch birds. In particular, at a significance level of 0.01, she would like to test the null hypothesis H₀: Average body mass of the Goldfinch birds is 0.5 oz, against the alternative claim that average body size is less than 0.5 oz.

      Assume that a random sample of 4 Goldfinch birds are going to be used to test the above hypothesis. Compute the largest value of the sample mean, that will allow the experimenter reject the null and prove the alternative at significance level 0.01. For this problem, make sure that your solution includes all steps of your computation. You will not earn points if you just use a formula

Answer: the largest value of the sample mean that will enable the experimenter to reject the null hypothesis is 0.4534 oz.

Step-by-step explanation:

Let us consider testing a hypothesis about a population mean with population standard deviation that is known,

Let the average body mass of the Goldfinch bird be μ^μ

Now our hypotheses are will be;

H₀ : μ = 0.5    and

H₁ : μ < 0.5

For the test, we shall use the z statistic which will be

z = (x"-μ)/(σ/√n)

Now given that;

σ = 0.04 and n = 4  

sample mean x" = ?

Now in order to be able to reject the null hypothesis, the p-value should be ≤ α = 0.01.

so to find the largest value of sample mean, we will use the maximum of p-value that is 0.01.

For a p-value = 0.01 at the lower tail (which is the percentile for z),

the corresponding z will be -2.33

so we substitute our values into our formula

z = (x"-μ)/(σ/√n)

z(σ/√n) = (x"-μ)

-2.33(0.04/√n) = (x"-0.5)

-0.0466 = x" - 0.5

x" = 0.5 - 0.0466

x"= 0.4534

therefore the largest value of the sample mean that will enable the experimenter to reject the null hypothesis is 0.4534 oz.

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Mimi started a pattern whit 5 and used the rule add 10
guajiro [1.7K]

Answer:

This seems like an incomplete question...

Step-by-step explanation:

The pattern would go 5...15...25...35...45...55... and so on.

Hope this is the answer you were looking for!!  :)

3 0
2 years ago
Find the output of the function y = x - 16 if the input is -28.<br> Help please
Serhud [2]

Answer:

-44

Step-by-step explanation:

If the input is -28, that means x = -28, so we plug -28 in for x and solve for y, which is the output:

y = x - 16

y = -28 - 16 = -44

The answer is thus -44.

5 0
3 years ago
Can someone please tell me how to find the value of x.
notsponge [240]

Answer:

2.547 = x

Step-by-step explanation:

tan 23 = x/6                  Use this equation to find the value of x

0.4245 = x/6                 Multiply both sides by 6

2.547 = x

7 0
2 years ago
The difference of twice a number and 4 is 16 .
german

Answer:

Let X be the number.

Twice the number would be 2x

The difference between the two, would be subtraction, so you would subtract 4 from 2x


The equation becomes 2x-4 = 16


Now solve for x:

2x-4 = 16

Add 4 to each side:

2x = 20

Divide both sides by 2:

x = 20/2

x = 10


The number is 10.


7 0
3 years ago
Read 2 more answers
Credit card companies lose money on cardholders who fail to pay their minimum payments. They use a variety of methods to encoura
Ivahew [28]

Answer with  explanation:

A x% confidence interval interprets that a person can be x% confident thatthe true mean lies in it.

Here, Credit card companies is using the collection agency to justify the cost of , the agency must collect an average of at least $200 per customer.

i.e. H_0:\mu \geq200,\ \ \ H_a:\mu

The 90% confidence interval on the mean collected amount was reported as ($190.25, $250.75) .

I recommend that we can be 90% sure that the true mean collected amount  lies in ($190.25, $250.75).

Also, $200 lies in it such that it is more far from $250.75 than $190.25, that means there are large chances of having an average is at least $200 per customer.

8 0
2 years ago
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