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Brut [27]
3 years ago
15

The t value for a 95% confidence interval estimation with 24 degrees of freedom is

Mathematics
1 answer:
Alborosie3 years ago
4 0

Answer: the value us 2.064

Step-by-step explanation:

To determine the t value, we would need the t distribution table. Since degree of freedom is known as 24, we would determine alpha/2

A 95% confidence level is 95/100 = 0.95

1 - alpha = 0.95

alpha = 1 - 0.95

alpha = 0.05

alpha/2 = 0.05/2 = 0.025

this is the area to the left. The area to the right 1 - 0.025 = 0.975

alpha/2 = 0.975

Looking at the t distribution table, the t value is

2.064

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Construct the 99% confidence interval estimate of the population proportion p if the sample size is n=900 and the number of succ
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Answer:

An 99% confidence interval  of the given proportion

(0.355 , 0.385)

Step-by-step explanation:

Given sample size n= 900

the number of successes in the sample is x=333

The proportion P = \frac{x}{n} = \frac{333}{900} = 0.37

            Q = 1-P =1 - 0.37 = 0.63

<u>Confidence interval</u>:-

99% of confidence interval zα = 2.93

(P - z_{\alpha } \sqrt{\frac{PQ}{n} }  , P + z_{\alpha } \sqrt{\frac{PQ}{n} })

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(0.37 - 0.015 , 0.37 + 0.015)

(0.355 , 0.385)

<u>Conclusion</u>:-

<u>An 99% Confidence interval (0.355 , 0.385)</u>

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What is the value of cos θ given that (-2, 9) is a point on the terminal side of θ?
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Answer:

a)  cosθ  = \frac{-2\sqrt{85} }{85}

Step-by-step explanation:

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

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