because the mean is not eual
To write 11 percent as a fraction, you divide the percent by 100 and simplify the fraction if possible. In this case, 11 percent is written as 11/100.
It is simply subtraction of the whole number from one hundredth of the percent: that is, A% - B = A/100 - B or (100*A - B)/100. It may be possible to simplify the answer.
No. No percentage higher than 100 percent is possible. Out of 100 occurrences it is impossible to obtain an answer of more than the totality of the occurrences.
Put them over 100. Reduce if possible. 27 percent = 27/100
Generally speaking an x% confidence interval has a margin of error of (100-x)%.
You can. For example, you can be 100% certain that the value when you roll a die will lie between 1 and 6. Or that the mean of 100 rolls will lie between 1 and 6. It is simply that a 100% CI has little use.
It will decrease.It will decrease
The formula for margin of error is (Z*)*(Standard Deviation))/(sqrt(N)), so as N increases, the margin of error decreases. Here N went from 100 to 5000, so N has increased by 4900. This means the margin of error decreases. Since the confidence interval is the mean plus or minus the margin of error, a smaller margin of error means that the confidence interval is narrower.
The confidence interval (this is the correct term) is a prediction made on an academic achievement test. The idea is that if the student was tested again the authors of the test feel that they can be confident that the student's score would fall within the range of the interval. Often, you'll see a number in percent such as 95%, and then a range of scores below it. Most academic achievement tests have a mean of 100, so the score will be somewhere in the range of the test. The number is based on a comparison between that student and the norm group that the test was tested on. The higher the percent of the confidence interval, the more reliable the test is.
No; it is not possible.
That is with a confidence interval of approximately 95% the "true mean" is within the interval of [336.10, 353.90] and that the sample mean (which is an estimate of the "true mean") is $350.00. SMALL-SAMPLE CONFIDENCE INTERVAL FOR A POPLATION MEAN, t-DISTRIBUTION 95% Confidence Interval = x-bar +/- (t-critical value) * s/SQRT(n) x-bar = SAMPLE MEAN [350] s = STANDARD DEVIATION [100] n = NUMBER OF SAMPLES [200] n - 1 = 199 df (DEGREES OF FREEDOM) t-critical value = (approx) 1.972 from "look-up Table for "two-sided interval" df = 200 [CLOSED df IN TABLE] 95% Confidence Interval: 350+/- 1.972 *100 / SQRT(200) = [336.10, 353.90] That is with a confidence interval of approximately 95% the "true mean" is within the interval of [336.10, 353.90] and that the sample mean (which is an estimate of the "true mean") is $350.00. c. ANSWER: A random selection of 1537 customers will provide 95% confidence for estimating the mean extended warranty price paid. Why??? CHOOSING THE SAMPLE SIZE n = [(z-critical value * s)/B]^2 z-critical value = 1.96 (associated with 95% confidence level) s = STANDARD DEVIATION [100.00] B = BOUND ON THE ERROR OF ESTIMATION [5.00] n = [(1.96 * 100.00)/5.00]^2 = 1537 (ROUNDED - UP) CONCLUSION: A random selection of 1537 customers will provide 95% confidence for estimating the mean extended warranty price paid.
because the mean is not eual
To write 11 percent as a fraction, you divide the percent by 100 and simplify the fraction if possible. In this case, 11 percent is written as 11/100.
Just put it over 100. 168 percent = 168/100
100% of anything is that anything. Just to confuse you even further, should that be possible, 409 percent of 100 is Yeah.... but im short the answer is 409.
yes