a. Patti and John each played 100 games of Solitaire on their ...

Question

a. Patti and John each played 100 games of Solitaire on their smartphones. Patti won 7 games and estimated her probability of winning Solitaire as 0.07. John won 6 games and estimated his probability as 0.06. Why do their estimates differ?

Accepted Answer

Hello. In this video, we are told that Carlos and Dana each inspected 200 manufactured widgets for defects. Carlos found 12 defect widgets, and Dana found 9 defective widgets. Which explanation best accounts for the differences between their estimated defect rates? So, if we want to determine what are the estimated defect rates for both Carlos and Dana, the first thing we want to do is we want to go ahead and find a sample proportion between them. So, the sample proportion is defined as the amount of widgets they found divided by the population size. Now, if we are determining the sample proportion for Carlos, so P had sub C, that's equal to the amount of defects he found, which is 12, divided by the total amount of defects, which is 200, and this is going to approximate to be 0.06. Now we will go ahead and find. The sample proportion for Dana, so P had sub D. Dana found 9 defective digits, so that's going to be 9 out of the 200 digits that they checked, and that's going to approximate to be 0.045. Now, what we can go ahead and do is we could go ahead and try to create A confidence interval between the digits of defects or the number of defects that Dana and Carlos found. So what we can go ahead and do is first determine the standard error. For both of them Starting with the standard error for Carlos, the standard error is going to be defined as the square root of 0.06 multiplied by 0.94, divided by 200, and that's going to give Carlos a standard error of 0.01679. Now, the standard error for Dana is going to be the square root of 0.045, multiplied by 0.955 divided by 200, and that is going to approximate to be 0.01466. Now, if we go back to the sample errors, and take a look at the difference of the sample errors, the difference is going to be 0.06 minus 0.045, and that is equal to 0.015. So what this means is that the difference is so significantly small and it's significantly small relative to the sample variance. So what we can go ahead and do is use this to create a 95% confidence interval for for each Carlos and Dana. Now the confidence interval boundaries are defined as the following. It is the sample proportion plus and minus 1.96 multiplied by the standard error with respect to each of the sample errors. So let's go ahead and create a confidence interval for Carlos. So for Carlos, his sample or his sample proportion was 0.06, and this is going to be plus or minus 1.96 multiplied by 0.01679, which is the standard error, and by simplifying the bounds of the confidence interval are going to be 0.0271 as the minimum value. And 0.0929 as the upper value. Now, if we repeat this process for Dana, Dana's bounds are going to be defined as 0.045 plus or minus 1.96, multiplied by their standard error. Which is 0.0737. And this is going to give us the approximate bound for the confidence interval with respect to data. That is going to be 0.0163 and 0.0737. Now if we take a look at the boundaries, the intervals overlap each other, so the observers. Difference It is plausible due to sampling variance. And so with that being said, if we go back to the given options, option B says that the difference is consistent with the sampling variability because the observed difference is small relative to the standard errors and their 95% confidence intervals overlap, and that's exactly what we saw when we found the confidence intervals with respect to their point estimates or the sample estimates. So with that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

a. Patti and John each played 100 games of Solitaire on their smartphones. Patti won 7 games and estimated her probability of winning Solitaire as 0.07. John won 6 games and estimated his probability as 0.06. Why do their estimates differ?

Key Concepts

Empirical Probability

Sample Size and Variability

Randomness and Chance

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