Confidence Intervals for the Mean

Chapter 23Confidence Intervals and Hypothesis Tests for a Population Mean ; t distributionst distributionst confidence intervals for a population mean Sample size required to estimate hypothesis tests for In the 2012-2013 NFL season Adrian Peterson of the Minn. Vikings rushed for 2,097 yards. The all-time single-season rushing record is 2,105 yards (Eric Dickerson 1984 LA Rams). Shown below are Petersons rushing yards in each game:84 60 86 102 88 79 153 123182 171 108 210 154 212 86 199We would like to estimate Adrian Petersons mean rushing ABILITY during the 2012-2013 season with a confidence interval.What is Adrian Petersons True Rushing ABILITY?

When we select simple random samples of size n, the sample means we find will vary from sample to sample. We can model the distribution of these sample means with a probability model that is

The Importance of the Central Limit Theorem

Since the sampling model for x is the normal model, when we standardize x we get the standard normal z

The sample standard deviation s provides an estimate of the population standard deviation sFor a sample of size n,the sample standard deviation s is:n 1 is the degrees of freedom.The value s/n is called the standard error of x , denoted SE(x). If is unknown, we probably dont know either.

Substitute s (sample standard deviation) for

Standardize using s for

ssss

ssss

Note quite correctNot knowing means using z is no longer correctSuppose that a Simple Random Sample of size n is drawn from a population whose distribution can be approximated by a N(, ) model. When s is known, the sampling model for the mean x is N(m, s/n). t-distributionsWhen s is estimated from the sample standard deviation s, the sampling model for the mean x follows a t distribution t(m, s/n) with degrees of freedom n 1.is the 1-sample t statistic

CONFIDENCE INTERVAL for

where:t = Critical value from t-distribution with n-1 degrees of freedom = Sample means = Sample standard deviationn = Sample sizeFor very small samples (n < 15), the data should follow a Normal model very closely.For moderate sample sizes (n between 15 and 40), t methods will work well as long as the data are unimodal and reasonably symmetric.For sample sizes larger than 40, t methods are safe to use unless the data are extremely skewed. If outliers are present, analyses can be performed twice, with the outliers and without.Confidence Interval Estimates

Very similar to z~N(0, 1)Sometimes called Students t distribution; Gossett, brewery employeeProperties:i)symmetric around 0 (like z)ii)degrees of freedom

t distributions

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Students t Distribution -3-2-10123Zt0123-1-2-3

Students t Distribution Figure 11.3, Page 372-3-2-10123Zt10123-1-2-3

Students t Distribution Figure 11.3, Page 372Degrees of Freedom

-3-2-10123Zt10123-1-2-3t7Students t Distribution Figure 11.3, Page 372

Degrees of Freedom

Degrees of Freedom13.07776.31412.70631.82163.65721.88562.92004.30276.96459.9250............101.37221.81252.22812.76383.1693............1001.29011.66041.98402.36422.62591.2821.64491.96002.32632.5758

0.800.900.950.980.9990% confidence interval; df = n-1 = 10t-Table

01.8125Students t Distribution P(t > 1.8125) = .05-1.8125.05.05.90t10P(t < -1.8125) = .05Conf.leveln = 30z = 1.64590%t = 1.6991z = 1.9695%t = 2.0452z = 2.3398%t = 2.4620z = 2.5899%t = 2.7564

Comparing t and z Critical ValuesIn the 2012-2013 NFL season Adrian Peterson of the Minn. Vikings rushed for 2,097 yards. The all-time single-season rushing record is 2,105 yards (Eric Dickerson 1984 LA Rams). Shown below are Petersons rushing yards in each game:84 60 86 102 88 79 153 123182 171 108 210 154 212 86 199Construct a 95% confidence interval for Petersons mean rushing ABILITY during the 2012-2013 season.Just 9 More Yards!

Because cardiac deaths increase after heavy snowfalls, a study was conducted to measure the cardiac demands of shoveling snow by handThe maximum heart rates for 10 adult males were recorded while shoveling snow. The sample mean and sample standard deviation wereFind a 90% CI for the population mean max. heart rate for those who shovel snow.Example

Solution

Determining Sample Size to Estimate Required Sample Size To Estimate a Population Mean If you desire a C% confidence interval for a population mean with an accuracy specified by you, how large does the sample size need to be?We will denote the accuracy by ME, which stands for Margin of Error.Example: Sample Size to Estimate a Population Mean Suppose we want to estimate the unknown mean height of male undergrad students at NC State with a confidence interval.We want to be 95% confident that our estimate is within .5 inch of How large does our sample size need to be?Confidence Interval for

Good news: we have an equationBad news:Need to know sWe dont know n so we dont know the degrees of freedom to find t*n-1

A Way Around this Problem: Use the Standard Normal

.95Confidence levelSampling distribution of x

Estimating sPreviously collected data or prior knowledge of the populationIf the population is normal or near-normal, then s can be conservatively estimated bys range699.7% of obs. Within 3 of the meanExample: sample size to estimate mean height of NCSU undergrad. male studentsWe want to be 95% confident that we are within .5 inch of , so ME = .5; z*=1.96Suppose previous data indicates that s is about 2 inches.n= [(1.96)(2)/(.5)]2 = 61.47We should sample 62 male students

Example: Sample Size to Estimate a Population Mean -TextbooksSuppose the financial aid office wants to estimate the mean NCSU semester textbook cost within ME=$25 with 98% confidence. How many students should be sampled? Previous data shows is about $85.

Example: Sample Size to Estimate a Population Mean -NFL footballsThe manufacturer of NFL footballs uses a machine to inflate new footballsThe mean inflation pressure is 13.5 psi, but uncontrollable factors cause the pressures of individual footballs to vary from 13.3 psi to 13.7 psi After throwing 6 interceptions in a game, Peyton Manning complains that the balls are not properly inflated.

The manufacturer wishes to estimate the mean inflation pressure to within .025 psi with a 99% confidence interval. How many footballs should be sampled?

Example: Sample Size to Estimate a Population Mean The manufacturer wishes to estimate the mean inflation pressure to within .025 pound with a 99% confidence interval. How may footballs should be sampled?99% confidence z* = 2.58; ME = .025 = ? Inflation pressures range from 13.3 to 13.7 psiSo range =13.7 13.3 = .4; range/6 = .4/6 = .067

12348. . .

Chapter 23Hypothesis Tests for a Population Mean 33

QTM1310/ Sharpe3325 pitchers with highest average fastball velocity:2007:2013:Was the ABILITY of the top 25 pitchers in 2013 to throw hard greater than the ABILITY of the top 25 pitchers in 2007 to throw hard?Are Major League Pitchers Throwing Harder? Average Fastball Velocity: 2007, 2013.

As in any hypothesis tests, a hypothesis test for requires a few steps:State the null and alternative hypotheses (H0 versus HA)Decide on a one-sided or two-sided testCalculate the test statistic t and determine its degrees of freedomFind the area under the t distribution with the t-table or technologyDetermine the P-value with technology (or find bounds on the P-value) and interpret the resultThe one-sample t-testStep 1:State the null and alternative hypotheses (H0 versus HA)Decide on a one-sided or two-sided testH0: m = m0 versus HA: m > m0 (1 tail test)H0: m = m0 versus HA: m < m0 (1 tail test)H0: m = m0 versus HA: m m0 (2 tail test)Step 2:

The one-sample t-test; hypotheses

We perform a hypothesis test with null hypothesisH0 : = 0 using the test statistic

where the standard error of is .

When the null hypothesis is true, the test statistic follows a t distribution with n-1 degrees of freedom. We use that model to obtain a P-value.The one-sample t-test; test statistic

38The one-sample t-test; P-ValuesRecall:The P-value is the probability, calculated assuming the null hypothesis H0 is true, of observing a value of the test statistic more extreme than the value we actually observed.

The calculation of the P-value depends on whether the hypothesis test is 1-tailed(that is, the alternative hypothesis isHA : < 0 or HA : > 0)or 2-tailed(that is, the alternative hypothesis is HA : 0).QTM1310/ Sharpe3839P-ValuesIf HA: > 0, then P-value=P(t > t0) Assume the value of the test statistic t is t0If HA: < 0, then P-value=P(t < t0)If HA: 0, then P-value=2P(t > |t0|)

QTM1310/ Sharpe3925 pitchers with highest average fastball velocity:2007:2013:Was the ABILITY of the top 25 pitchers in 2013 to throw hard greater than the ABILITY of the top 25 pitchers in 2007?Are Major League Pitchers Throwing Harder? Average Fastball Velocity: 2007, 2013.H0: = 95.92 HA: > 95.92

where is the average fastball velocity of the top 25 2013 pitchers

n = 25; df = 24Are Major League Pitchers Throwing Harder? Average Fastball Velocity: 2007, 2013.Reject H0: Since P-value < .05, there is sufficient evidence that top 25 pitchers in 2013 on average throw harder0.008

2. 58H0: = 95.92 HA: > 95.92

t, 24 df

Conf. Level0.10.30.50.70.80.90.950.980.99Two Tail0.90.70.50.30.20.10.050.020.01One Tail0.450.350.250.150.10.050.0250.010.005dfValues of t240.12700.39000.68481.05931.31781.71092.06392.49222.7969P-value = .008

42Are Major League Pitchers Throwing Harder? Average Fastball Velocity: 2007, 2013.2.4922 < t = 2.58 < 2.7969; thus 0.01 < p < 0.005.

0.0082. 58t, 24 dfA popcorn maker wants a combination of microwave time and power that delivers high-quality popped corn with less than 10% unpopped kernels, on average. After testing, the research department determines that power 9 at 4 minutes is optimum. The company president tests 8 bags in his office microwave and finds the following percentages of unpopped kernels: 7, 13.2, 10, 6, 7.8, 2.8, 2.2, 5.2.Do the data provide evidence that the mean percentage of unpopped kernels is less than 10%?Microwave PopcornH0: = 10HA: < 10where is true unknown mean percentage of unpopped kernels

n = 8; df = 7Microwave PopcornReject H0: there is sufficient evidence that true mean percentage of unpopped kernels is less than 10%.020

H0: = 10 HA: < 10-2. 51

t, 7 df

Exact P-value = .02Conf. Level0.10.30.50.70.80.90.950.980.99Two Tail0.90.70.50.30.20.10.050.020.01One Tail0.450.350.250.150.10.050.0250.010.005dfValues of t70.13030.40150.71111.11921.41491.89462.36462.99803.49952.3646 < |t| = 2.51 < 2.9980 so .01 < P-value < .025