Over the first 20 20 days of the semester, one student is late to class on 6 6 days. Find the margin of error for a 98 % 98\% confidence interval for the true proportion of time this student is late.

Over the 1st 20 days of the semester, one student is late on 6 of those days. We're asked to find here the margin of error for a 98% confidence interval for the true proportion of time that this student is late. Now for this particular problem, all we're asked to do so far is find the margin of error, which we can do using this equation here. Now we don't actually have to construct our confidence interval, so we're only going up to step 4 here, finding that margin of error without actually finding the upper and lower bounds of our interval. So let's go ahead and start from step 1. We first wanna verify that the number of successes and failures are both greater than or equal to 5. Now in this case, this student is late to class on 6 days. Now whether or not you consider it a success to be late to class, that is the context of this problem. So on 6 days, those are number of successes that is greater than or equal to 5. That means on 14 of those days, they were on time. Those are considered our failures in this case. That is also greater than or equal to 5, so we are good to go there. We can move on to step 2, finding our critical z value. Now in finding our critical z value, this is dependent on that confidence level. So our confidence level here is 98 percent. So finding alpha over 2 here, 1 minus 0.98 over 2, that gives us here 0.01, which using a z table or a calculator, this will end up telling us that our critical z value, z alpha over 2, is equal to 2.326. Now our next step here is to find p hat and p hat is equal to x over n. So in this case, p hat is going to be equal to the number of days that the student was late, that's x, over those total number of days, that is 20 in this case. So 6 over 20, that gives us 0.3. Now that we have p hat, we can go ahead and actually find our margin of error using this equation here. So we know that our margin of error, e, is equal to z alpha over 2 times the square root of p hat times 1 minus p hat over n. So plugging in all of those values here, that's going to be 2.326 times the square root of 0.3 times 1 minus 0.3, that's going to be 0.7 over n, which is 20. Now working out this value will end up being 0.238 for our margin of error, and that's all for now because all we were asked to do was find the margin of error. Once we're asked to find the confidence interval, we can use that margin of error along with our point estimate p hat to construct our confidence interval. Feel free to let us know if you have any questions, and I'll see you in the next one.

8. Sampling Distributions & Confidence Intervals: Proportion

Confidence Intervals for Population Proportion

Statistics

8. Sampling Distributions & Confidence Intervals: Proportion

Confidence Intervals for Population Proportion

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Multiple Choice

Over the first $20$ days of the semester, one student is late to class on $6$ days. Find the margin of error for a $98 percent sign$ confidence interval for the true proportion of time this student is late.

$E=0.3$

$E=0.238$

$E=0.169$

$E=0.062$

Verified step by step guidance

Step 1: Begin by identifying the sample proportion (p̂) of days the student is late. This is calculated by dividing the number of days late (66) by the total number of days (2020).

Step 2: Calculate the standard error (SE) of the sample proportion using the formula: SE = sqrt((p̂ * (1 - p̂)) / n), where n is the total number of days (2020).

Step 3: Determine the z-score for a 98% confidence interval. This can be found using a standard normal distribution table or calculator, which typically gives a z-score of approximately 2.33 for 98% confidence.

Step 4: Calculate the margin of error (E) using the formula: E = z * SE, where z is the z-score from Step 3 and SE is the standard error from Step 2.

Step 5: Interpret the margin of error in the context of the problem. This value represents the range within which the true proportion of days the student is late is expected to fall, with 98% confidence.