confidence intervals
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Confidence Intervals. Means. Remember: In sampling distribution of means…. Approximately normal if n is large ( n 30) – CLT The population is normal. So how will I find a confidence intervals for a mean? How is it different from a proportion?. Formula for a z-confidence interval:. - PowerPoint PPT PresentationTRANSCRIPT
Confidence Intervals
Means
Remember: In sampling distribution of means…..
Approximately normal if n is large ( n 30) – CLTThe population is normal
x
x n
So how will I find a confidence intervals for a mean? How is it different from a proportion?
Formula for a z-confidence interval:
2x z
n
Suppose you work for a consumer advocate agency & want to find the mean repair cost of a washing machine. You randomly select 40 repair costs and find the mean to be $100. The standard deviation is $17.50. construct a 95% interval for the population mean.
Find the minimum required sample size if you want to be 95% confident that the sample mean is within 2 units of the population mean if the population standard deviation is 4.8.
But what if the sample size is small or the population standard deviation is unknown?
We use the t-distribution if the population standard deviation is unknown.
It’s bell shaped – centered at 0. Each t-distribution is more spread out than
the z-distribution (normal). As the sample size increases, the spread
decreases and actually approaches the normal distribution.
Based on degrees of freedom (df = n-1)
Degrees of freedom: (n-1)
Suppose A + B + C + D = 18
Thus there are 3 degrees of freedom
Free to be anything
The t- distribution compared to the normal distribution http://www.nku.edu/~longa/stats/taryk/TDist
.html
In normal sampling but with t-distribution
xz
n
x
tsn
Reading the t-chart
The Raman Arches is an Italian restaurant. The manager wants to estimate the average amount a customer spends on lunch. A random sample of 115 customers’ lunch tabs gave a mean of $9.74 with a standard deviation of $2.93. Find and interpret a 90% confidence interval for the average amount spent on lunch by all customers.
A sample of 20 students had a test average of 100 with a standard deviation of 4.2 points. Find and interpret a 90% confidence interval for the average test score for all students on this test.
A random sample of large tents listed in Consumer Reports: Special Outdoor Issue gave the following prices. Find a 90% confidence interval for the mean price of all such tents.
115 110
140 135
80 110
150 210
250 120
230 130
Homework - Worksheet