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Masja [62]
4 years ago
15

A confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the con

fidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A 99%99% confidence interval for μμ: 100 to 136 (a) H0: μ=88H0: μ=88 vs Ha: μ≠88Ha: μ≠88 Conclusion: Choose the answer from the menu in accordance to item (a) of the question statement H0H0 Significance level: Choose the answer from the menu in accordance to item (a) of the question statement (b) H0: μ=102H0: μ=102 vs Ha: μ≠102Ha: μ≠102 Conclusion: Choose the answer from the menu in accordance to item (b) of the question statement H0H0 Significance level: Choose the answer from the menu in accordance to item (b) of the question statement
Mathematics
1 answer:
statuscvo [17]4 years ago
7 0

Answer:

Step-by-step explanation:

Given is the results of a hypothesis testing for sample mean with population mean.

H_0; \bar x = 88\\H_a: \bar x \neq 88

(Two tailed test at 1% significance level)

Reason:Two tailed because confidence interval is (100,136) with mean =118

Significance level = 100-confidence level

Given that confidence interval is (100,136)

i.e. we can be 99% confident that if samples for large sizes are drawn from this population will have mean within this interval

In this case sample mean 88 does not fall within this interval

Hence our null hypothesis has to be rejected

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evablogger [386]

Answer:

Edger needs 15 cookies and 5 brownies.

Step-By-Step:

6 + 6 + 3 = 15

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3 years ago
Find a function where f(0)=2 and f(1)=2
xenn [34]

Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

f(x) = 2 is a valid solution.

Want something more fun? Why not a parabola? f(x)= ax^2+bx+c.

At this point you have three parameters to play with, and from the fact that f(0)=2 we can already fix one of them, in particular c=2. At this point I would recommend picking an easy value for one of the two, let's say a= 1 (or even a=-1, it will just flip everything upside down) and find out b accordingly:f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1

Our function becomes

f(x) = x^2-x+2

Notice that it works even by switching sign in the first two terms: f(x) = -x^2+x+2

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: f(x) = A cos (kx)

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need A= 2, and at that point the first condition is guaranteed; using the second to find k we get 2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi

f(x) = 2cos(2\pi x)

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

f(x)= 2+sin(2\pi x)

Sky is the limit.

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Can someone please help me with this???
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