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8.2 Estimating μ When σ is Unknown

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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.

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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.

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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?

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The t Distribution has a shape similar to that of the the Normal Distribution A Normal distribution A “ t ” distribution

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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:

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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.

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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

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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 t 0.90 = 1.895.

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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

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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.)

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