Aim: What are confidence intervals for means that have unknown standard deviations and sample sizes...

Post on 17-Jan-2018

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How T distribution differs from standard normal distribution The variance is greater than 1 The t distribution is actually a family of curves based on the concept of degrees of freedom, which is related to sample size As the sample size increases, the t distribution approaches the standard normal distribution

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Aim: What are confidence intervals for means that have unknown

standard deviations and sample sizes less than 30?

Quiz Friday

T distribution

• Used when sample size is less than 30 and the variable is normally or approximately normally distributed

• Important Characteristics:– It is a bell-shape– It is symmetric about the mean– The mean, median and mode are equal to 0 and

are located at the center of the distribution– The curve never touches the x axis

How T distribution differs from standard normal distribution

• The variance is greater than 1• The t distribution is actually a family of curves

based on the concept of degrees of freedom, which is related to sample size

• As the sample size increases, the t distribution approaches the standard normal distribution

Degrees of freedom

• Degrees of freedom: the number of values that are free to vary after a sample statistics has been computed– Tells the researcher which specific curve to sue

when a distribution consists of a family of curves

d.f. = n -1

The t family of Curves

Formula for a specific confidence interval for the mean

• Use this when standard deviation is unknown and n < 30

• The values of tα/2 are found in table F. The top row of table F, labeled confidence intervals is used to get these values

2 2

s sX t X tn n

Using the F Table

Example

• Ten randomly selected automobiles were stopped and the tread depth of eh right front tire was measured. The mean was 0.32 inch, and the standard deviation was 0.08 inch. Find the 95% confidence interval of the mean depth. Assume the variable is approximately normally distributed.

Solution• Since σ is unknown and s must replace it, the t

distribution must be used for 95% confidence interval. Hence the d.f. = 9, tα/2 =2.262 (from table F).

2 2

.08 .08.32 2.262 .32 2.26210 10

.32 0.057 .32 .057

.26 .38

s sX t X tn n

Class Work1. Find the values for each

1. tα/2 and n = 18 for the 99% confidence interval for the mean2. tα/2 and n =20 for the 95% confidence interval for the mean

2. A recent study of 25 students showed that they spent an average of $18.53 for gasoline per week. The standard deviation of the sample was $3.00. Find the 95% confidence interval of the true mean.

3. The number of unhealthy days based on the Air Quality Index for a random sample of metropolitan areas is shown. Construct a 98% confidence interval based on the data.

61 12 6 40 27 3893 5 13 40