8.2 Estimating μ When σ is Unknown. What if it is impossible or impractical to use a large sample?...
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Transcript of 8.2 Estimating μ When σ is Unknown. What if it is impossible or impractical to use a large sample?...
8.2
Estimating μ When σ is Unknown
What if it is impossible or impractical to use a large sample?
Apply the Student’s t distribution:
It is important to note that the t distribution uses a degrees of freedom (d.f.) = n – 1
The shape of the t distribution depends only on the sample size, n, if the basic variable x has a normal distribution.
When using the t distribution, we will assume that the x distribution is normal.
Student’s t Distributions
• In order to use the normal distribution to find confidence intervals for a population mean μ we need to know the value of σ. However, much of the time, we don’t know either. In such cases, we use the sample standard deviation s to approximate σ.
• When we use s to approximate σ, the sampling distribution for x bar follows a distribution known as Student’s t distribution.
More about t distribution
• The graph is always symmetrical about its mean, which (as with z dist) is 0.
• t distribution is similar to normal z distribution except it has somewhat thicker tails. (as seen on next slide)
• As the degrees of freedom increase, the t distribution approaches the standard normal distribution.– As n increases… what happens to the distribution of
any sample?
The t Distribution has a shape similar to that of the the Normal Distribution
A Normal distribution
A “t” distribution
Maximal Margin of Error, E
Similar to what we saw in the previous section, we can determine the confidence interval by
using the margin of error:
Example
A company has a new process for manufacturing large artificial sapphires. In a trial run, 37 sapphires are
produced. The mean weight for these 37 gems is 6.75 carats and σ = 0.33 carats. Let μ be the mean weight for
distribution of all.
Why is the t distribution the preferred method here?• Find E for a 95% confidence interval. • Then find the confidence interval.• Interpret the confidence interval in the context of the
problem.
The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of
2.3 ounces.
Construct a 90% confidence interval for the mean weight of the population of fish in the lake.
Another Example
Mean = 15.7 ounces Standard deviation = 2.3 ounces.
• n = 8, so d.f. = n – 1 = 7• For c = 0.90, Table 6 in Appendix II gives
t0.90 = 1.895.
Mean = 15.7 ounces Standard deviation = 2.3 ounces.
E = 1.54
The 90% confidence interval is:
15.7 - 1.54 < < 15.7 + 1.54
14.16 < < 17.24
ExEx
Summary
Examine problem Statement
σ is known
Use normal distribution with margin of error
σ is unknown
Use Student t’s distribution with
margin of error (don’t forget about d.f.)