An experimenter would like to construct a 99% two-sided t-interval, with a length at almost 0.2 ohms, for the average resistance of a segment of copper cable of a certain length. if the experimenter feels that the standard deviation of such resistances is no larger than 0.15 ohms, what sample size would you recommend? assume that an appropriate t-value is 3.0.a. aprox 10b. aprox 20c. aprox 30d. aprox 40e. aprox 50

Aprox 20

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

represent the sample mean for the sample  

population mean

s represent the sample standard deviation

n represent the sample size (variable of interest)

Confidence =99% or 0.99

The margin of error is given by this formula:

   (1)

And on this case we have that ME =0.2/2=0.1 ohms, since the lenght of the confidence interval is L=2ME and we are interested in order to find the value of n, if we solve n from equation (4) we got:

  (2)

The critical value for 99% of confidence interval is provided, , replacing into formula (2) we got:

So the answer for this case would be n=21 rounded up to the nearest integer

And with the options provided the best answer is:

Aprox 20

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads

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