AP Statistics

If our data comes from a simple random sample (SRS) and the sample size is sufficiently large, then we know that the sampling distribution of the sample means is approximately normal with mean μ and standard deviation .

The spread of the sampling distribution depends on n and σ. σ is generally unknown and must be estimated.

NOW…THEORY ASIDE AND ONTO PRACTICE !

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n

SRS – size nNormal distribution of a populationμ and σ are unknownTo estimate σ – use “S” in its place

Then the standard error of the sample mean is

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sn

The z statistic has N (0,1)When s is substituted the distribution is no longer

normal

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xz

n

The t statistic is used when we don’t know the standard deviation of the population, and instead we use the standard deviation of the sample distribution as an estimation.

The t statistic has n-1 degrees of freedom (df).

/xts n

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Interpret the t statistic in the same way as the z statistic

There is a different distribution for every sample size.

The t statistic has n-1 degrees of freedom.

Write t (k) to represent the t distribution with k degrees of freedom.

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Density curves for the t distribution are similar to the normal curve (symmetrical and bell shaped)

The spread is greater and there is more probability in the tails and less in the center.

Using s introduces more variability than sigma.

As d.f. increase, t(k) gets more normal

In statistical tests of significance, we still have H0 and Ha.

We need to provide the mu in the calculation of the t statistic.

Looking at the t table is fundamentally different than the z table.

/xts n

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Assume SRS size n with population mean μ Confidence interval will be correct for normal

populations and approx. correct for large n.

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estimate t * (SE estimate)

sCI x t * ( )n

(1 C)t* f or t(n-1)2

Let’s suppose that Mr. Young has been told that he should mop the floor by 1:25 p.m. each day.

We collect 12 sample times with an average of 27.58 minutes after 1 p.m. and with a standard deviation of 3.848 minutes.

Find a 95% confidence interval for Mr. Young’s mopping times.

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x 27.58mins 3.848n 12df 11CL :95%

From table C:t* = 2.201

3.848CI 27.58 2.20112

CI : (25.135, 30.025)

Step 1: Population of interest:

◦ Mr. Young’s mopping time Parameter of interest:

◦ average time of arrival to mop

Hypotheses◦ H0: µ=25 min past 1:00◦ Ha: µ>25 min past 1:00

/xts n

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We are using 1 sample t-test? Bias?

◦ SRS not stated. Proceed with caution. Independence?

◦ Population size is at least 10 times the sample size?

◦ We assume that Mr. Young has mopped on a lot of days

Normality?◦ Big sample size (> 30). No◦ Sample is somewhat normal because

the sample distribution is single peaked, no obvious outliers.

/xts n

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Calculate the test statistic, and calculate the p-value from Table C

27.58 253.848 / 122.322

( 2.322) is between .025 and .02

t

P t

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Is the t-value of 2.322 statistically significant at the 5% level? At the 1% level?

Does this test provide strong evidence that Mr. Young arrives on time to complete his mopping?

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Try this exercise on your calculator using:STAT TESTS TintervalSTAT TESTS T-Test

Wednesday: 11.6 – 11.11 Thursday: 11.13 – 11.20 Friday: T-Test Worksheet

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