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### The t Distribution

Warm-up

- Suppose that we check for clarity in 50 locations in Lake Tahoe and discover that the average depth of clarity of the lake is 14 feet. Suppose that we know that the standard deviation for the entire lake's depth is 2 feet and it is normally distributed. What can we conclude about the average clarity of the lake with a 95% confidence level?
- Answer: (13.45,14.55)

Ch.10

Normal Distribution and t-distributions.

- When we know the population standard deviation (σ). We use the normal curve
- When we do not know the population standard deviation. We use what is called the t-distribution and sub in s for the population standard deviation (σ).
- There is a different t-distribution for each sample size n. The larger the sample size the closer it looks like the normal curve.

How does a t-distribution differ from a z-distribution?

- Applet
- http://www.nku.edu/~longa/stats/taryk/TDist.html
- Compare the distributions

Difference between a t-distribution and a z-distribution

- Similar in shape
- Spread of t-distribution is a bit greater than that of the standard Normal distribution.
- Tails of the t-distribution have more area
- As the degree of freedom increases the density curve approaches the normal curve.

t-distributions

- T-distributions have what is called a degree of freedom df=n-1
- A level C confidence interval for the mean of the population is
- We have t* instead of z* and s instead of σ

How do you find t*?

- We use table C to determine the correct value of t* for a given confidence interval.
- Find the value of t* given
- 1) 95% confidence level and n=6
- t*=2.571
- 2) 99% confidence level and n=23
- t*=2.819
- 3) 90% confidence level and n=45
- Hint use the greatest df available that is less than your desired df
- t*=1.684

How to find t* in the calculator? book)

- t*= invT( (1+c)/2, df)
- This can be used for any confidence Level

Another Example book)

- Suppose that we conduct a survey of 19 millionaires to find out what percent of their income the average millionaire donates to charity. We discover that the mean percent is 15 with a standard deviation of 5 percent. Find a 95% confidence interval for the mean percent. Assume that the distribution of all charity percents is approximately normal.
- Answer: We can conclude with 95% confidence that the millionaires donate between 12.6% and 17.4% of their income to charity.

Example of finding a confidence interval with unknown population standard deviation

- Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given below. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.
- 8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9
- Answer (7.31, 9.15)

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