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Confidence intervals for means

Confidence intervals for means. Sec 9.1 Continued. Finding confidence intervals for means:. We can find confidence intervals for a populations mean. While this will not be our preferred tool, it can be very effective in many situations. We can find confidence intervals with,

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Confidence intervals for means

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  1. Confidence intervals for means Sec 9.1 Continued

  2. Finding confidence intervals for means: • We can find confidence intervals for a populations mean. While this will not be our preferred tool, it can be very effective in many situations. • We can find confidence intervals with, • Where the standard error is, Z-score for confidence level Sample proportion Standard Error

  3. Example: • You sample 100 people and find those 100 people drink an average of 6.4 glasses of water a day, with a standard deviation of 2.5 glasses. Find a 95% confidence interval for the average number of glasses of water a person drinks in a day. • Notice we do not know σ, we will use s, the sample standard deviation in its place. When n is large this assumption holds well enough for most purposes. • So in practice standard error is,

  4. Questions… • How ‘large’ do we need to be to be close enough? • What if ‘n’ is not ‘large’?? • What if ‘close enough’ is not ‘good enough’???

  5. Answer… • The t-distribution!!! • The t-distribution is like the z-distribution that gives us our z-scores but it takes into account a few more things. • First, it can handle smaller sample sizes. • Second, it is adjusted to account for the error caused by using the sample standard deviation instead of the population’s standard deviation. • It works by being slightly wider and thicker then the z-distribution and having different values depending on sample size.

  6. A distribution that gives us extra space. , or t-distribution for n = ∞ z-distribution t-distribution t-distribution for smaller n

  7. How do we find a t-score? • To find a t-score we need to know two things. • The Sample Size, • The sample size give us what is called the “degrees of freedom” for the sample. df = n - 1 • The confidence level we want, • And the size of the ‘tails’ left outside those intervals Tail = (1 – CL) / 2

  8. Example, • You want to find an average for the expenditure of fuel during the launch of an orbital vehicle. After seven trials you find an average of 467 pounds of fuel with a standard deviation of 43.2 pounds. Find a 95% confidence interval for the pounds of fuel on an average launch. Instead of; We need to use;

  9. Example, • You want to find an average for the expenditure of fuel during the launch of an orbital vehicle. After seven trials you find an average of 467 pounds of fuel with a standard deviation of 43.2 pounds. Find a 95% confidence interval for the pounds of fuel on an average launch. We need to use; What is the degree of freedom? n – 1 = 7 – 1 = 6 df = 6 What is the size of the ‘tail’? 1 - .95 = 0.05 0.05 / 2 = 0.025

  10. Finding the t-score. • Given df = 6 and CL of 95% find the t-score. SO, becomes,

  11. Example, • You want to find an average for the expenditure of fuel during the launch of an orbital vehicle. After seven trials you find an average of 467 pounds of fuel with a standard deviation of 43.2 pounds. Find a 95% confidence interval for the pounds of fuel on an average launch. We need to use;

  12. Comparing populations; • When we want to know if two populations differ we often use confidence intervals. • To make a strong claim that a value is different for two separate groups their respective confidence intervals should NOT overlap. • Conversely to show two groups are the same in regards to some statistical values you can show that their confidence intervals overlap even for large sample sizes.

  13. Example; Note: • A company wants to test the effects of adding a new compound to it’s cholesterol drug. They give the old drug to 50 people and the new drug to 50 people. • The first group has an average of a 13.2 point drop in their cholesterol, with a standard deviation of 1.2 points. • The second group has an average drop of 14.1 with a standard deviation of 2.3 points. • From this study what can be said about the effects of the new compound? • Construct a 95% CI for each group. (using t-scores)

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