Introduction to Confidence Intervals (page 248)
In chapter 7 we discussed how to make inferences about a population parameter based on a sample statistic. While this can be useful, it has severe limitations. In Chapter 8, we expand our toolbox to include Confidence Intervals. Instead of basing our inference on a single value, a point estimate, a Confidence Interval provides a range of values, an interval, which – at a certain level of confidence (90%, 95%, etc.) – contains the true population parameter. Having a range of values to make inferences about the population provides much more room for accuracy than making an inference off of only one value.
When we worked with probabilities based on sample means, we learned that there is
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Assume that the population standard deviation is fairly stable at 1.8 hours.
Calculate the 95% confidence interval for the population mean weekday sleep time of all adult residents of this Midwestern town.
95% = 1.96 6.4 +(-) 1.96 x (1.8/sqrt80) = [6.01, 6.79]
Can we conclude with 95% confidence that the mean sleep time of all adult residents in this Midwestern town is not 7 hours?
Yes, we can conclude with 95% confidence that the mean sleep in this Midwestern town is not 7 hours because the value 7 does not fall within the confidence interval.
Confidence Interval for the Population Mean when Sigma is Unknown
While it is possible that we could know enough about our population to make an assumption about what the population standard deviation is, it is much more likely that if we do not know the population mean, then we do not know the population standard deviation. In this case, we can’t use the standard normal distribution, and we use a different distribution, the Student’s t distribution. Instead of , we use s, the sample standard deviation.
The formula is: x±ta/2, df sn .
/2 is still defined the same way, and df is degrees of freedom, calculated as n-1, where n is the sample size. Degrees of freedom determine the extent of the broadness of the tails of the distribution; the fewer the degrees of freedom, the broader the tails.
Solve the following problems:
Find ta/2, df for the following confidence levels: T-SCORE
Hypothesis: Sleeping for more than 8 hours has an effect on the exam score of college students. Null Hypothesis: Sleeping for more than 8 hours does not have an effect on the exam score of college students. The first threat to international
We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:
2. In order to determine the average amount spent in November on Amazon.com a random sample of 144 Amazon accounts were selected. The sample mean amount spent in November was $250 with a standard deviation of $25. Assuming that the population standard deviation is unknown, what is a 95% confidence interval for the population mean amount spent on Amazon.com in November?
(21) You took a sample of size 21 from a normal distribution with a known standard deviation, . In order to find a 90% confidence interval for the mean, You need to find.
5. Give the standard deviation for the mean and median column. Compare these and be sure to identify which has the least variability?
Compute the population mean margin of error for a 90% confidence interval when sigma is 7 and the sample size is 36.
7. Give and interpret the 95% confidence interval for the hours of sleep a student gets.
σA = 0.3 × (0.07)2 + 0.4 × (0.06)2 + 0.3 × (0.08)2 − (0.021)2 = 0.004389,
The amount of time it takes to recover physiologically from a certain kind of sudden noise is found to be normally distributed with a mean of 80 seconds and a standard deviation of 10 seconds. Using the 50%–34%–14% figures, approximately what percentage of scores (on time to recover) will be:
The customers in this case study have complained that the bottling company provides less than the advertised sixteen ounces of product. They need to determine if there is enough evidence to conclude the soda bottles do not contain sixteen ounces. The sample size of sodas is 30 and has a mean of 14.9. The standard deviation is found to be 0.55. With these calculations and a confidence level of 95%, the confidence interval would be 0.2. There is a 95% certainty that the true population mean falls within the range of 14.7 to 15.1.
Fry Brothers heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. Assume the population standard deviation for Larry Clark is 1.05 calls per day and 1.23 calls per day for George Murnen. A random sample of 40 days last year showed that Larry Clark made an average of
The last question asks if living on or off campus affects the amount of sleep a Truman student receives on average per day.
3.Find a 95% confidence interval for the mean tenure(in months) of all employees who leave D&Y within 3 years of being Hired. Why is it not possible with the given data to find a confidence interval for the
The 95% confidence interval for the population mean is 66,438 to 80,241. This means that there is a 95% confidence that this interval has the population mean.