Posted by Anonymous on Thursday, May 9, 2013 at 1:16pm.
1. 99% = mean ± 2.575 SEm
SEm = SD/√n
2. ± 1.96 SEm
3. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) to get the value for the Z score.
4. Same table, but look for .10.
1. A study of 40 people found that they could do on the average15 pull ups with a standard deviation of .6. Find the 99% confidence interval for the mean of the population.
Confidence Interval:
±0.24
Range for the true population mean:
14.76 to 15.24
2. A study of 50 pizza delivery workers found that they could make 6 deliveries per hour with a standard deviation of 1. Find the 95% confidence interval of the mean for all pizza delivery workers.
Confidence Interval:
±0.28
Range for the true population mean:
5.72 to 6.28
3. A random sample of 50 households showed that the average number of tv's in each household is 2.3 with a standard deviation of .4. Find the 90% confidence level for the average number of tv's in every household in the population.
Confidence Interval:
±0.09
Range for the true population mean:
2.21 to 2.39
4. A researcher revealed that the average number of people who have a driver's license out of a sample of 100 people is 1.5 with a standard deviation of .3. Find the 80% confidence interval for the mean of the population.
Confidence Interval:
±0.04
Range for the true population mean:
1.46 to 1.54
5. The following random sample was selected : 4, 6, 3, 5, 9, 3. Find the 95% confidence interval for the mean of the population.
Confidence Interval:
±0.24
Range for the true population mean:
1.76 to 2.24
6. In a sample of 35 high school seniors, 14 of them are attending college in the fall. Find the 95% confidence interval for the true proportion of high school seniors that will attend college in the fall from the population.
0.14 +/- (1.96)(√0.14)/(0.86)/35)
0.16436428065
0.1643 - .5/1000 = 0.1638 -.0005 = 0.1633
0.1643 + .5/1000 = 0.1648 + .0005 = 0.1653
16.33%-16.53%
7. In a sample of 200 people, 76 people would rather work out at home than in a gym. Find the 99% confidence interval for the true proportion of people who would rather work out at home than in a gym for the entire population.
0.76 +/- (2.575) ((√0.76)/(-0.24)/200)
0.71323264675
0.713 - .5/1000 = 0.7125 -.0005 = 0.712
0.713 + .5/1000 = 0.7135 + .0005 = 0.714
71.2%-71.4%
8. A study found that out of 300 people 60% of them prefer to eat hamburgers rather than hot dogs. Fin the 95% confidence interval for the true proportion of people who prefer to eat hamburgers rather than hot dogs in the entire population.