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A scientist measured the speed of light. His values are in​ km/sec and have​ 299,000 subtracted...

A scientist measured the speed of light. His values are in​ km/sec and have​ 299,000 subtracted from them. He reported the results of 29 trials with a mean of and a standard deviation of 104.77. ​

a) Find a ​98% confidence interval for the true speed of light from these statistics. ​

b) State in words what this interval means. Keep in mind that the speed of light is a physical constant​ that, as far as we​ know, has a value that is true throughout the universe.

​c) What assumptions must you make in order to use your​ method? ​

a) A ​98% confidence interval for the true speed of light is ​(______________km/sec,_______,(Round to two decimal places as​ needed.) ​

b) In​ words, what does the 98​% confidence interval​ mean?

A. The confidence interval contains the true speed of light ​98% of the time.

B. For all​ samples, ​98% of them will have a mean speed of light that falls within the confidence interval.

C. With ​98% ​confidence, based on these​ data, the speed of light is between the lower and upper bounds of the confidence interval.

D. Any measurement of the speed of light will fall within this interval 98% of the time

c)What assumptions must you make in order to use your method? Select all that apply.

A. The data come from a distribution that is nearly uniform.

B. The data come from a distribution that is nearly normal.

C. The measurements arise from a random sample or suitability randomized experiment.

D. The measurements are independent.  

E. The sample is drawn from a large population.

D. Any measurement of the speed of light will fall within this interval ​% of the time. ​

c) What assumptions must you make in order to use your​ method? Select all that apply.

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Answer #1

Solution:

Given data

n=29,\bar{x}= 756.24, s = 104.77

(a)

Degree of freedom: df =n-1=28

For 98% confidence interval, t -critical value is 2.4851. Therefore required confidence interval is

\bar{x}\pm t_{critical}\frac{s}{\sqrt{n}}=756.24\pm 2.4851\cdot \frac{104.77}{\sqrt{29}}=756.24\pm 48.35= ( 707.89, 804.59 )

Therefore, a 98% confidence interval for the mean is ( 707.89, 804.59).

b) Option (C) With ​98% ​confidence, based on these​ data, the speed of light is between the lower and upper bounds of the confidence interval.

c) The assumptions are

B. The data come from a distribution that is nearly normal.

D. The measurements are independent.  

E. The sample is drawn from a large population.

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