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t distributions t confidence intervals for a population mean Sample size required to estimate hypothesis tests for

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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 Peterson’s rushing yards in each game: 84 60 86 102 88 79 153 123 182 171 108 210 154 212 86 199 We would like to estimate Adrian Peterson’s mean rushing ABILITY during the 2012-2013 season with a confidence interval.

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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

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The sample standard deviation s provides an estimate of the population standard deviation For 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).

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Substitute s (sample standard deviation) for ssss s ss s Note quite correct Not knowing means using z is no longer correct

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Suppose that a Simple Random Sample of size n is drawn from a population whose distribution can be approximated by a N(µ, σ) model. When is known, the sampling model for the mean x is N( /√n). When is estimated from the sample standard deviation s, the sampling model for the mean x follows a t distribution t( , s/√n) with degrees of freedom n − 1. is the 1-sample t statistic

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CONFIDENCE INTERVAL for where: t = Critical value from t-distribution with n-1 degrees of freedom = Sample mean s = Sample standard deviation n = Sample size For 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.

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Very similar to z~N(0, 1) Sometimes called Student’s t distribution; Gossett, brewery employee Properties: i)symmetric around 0 (like z) ii)degrees of freedom

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-3-20123 Z 0123 -2-3

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-20123 Z t 0123 -2-3 Figure 11.3, Page 372

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-3-20123 Z t1t1 0123 -2-3 Figure 11.3, Page 372 Degrees of Freedom

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-3-20123 Z t1t1 0123 -2-3 t7t7 Figure 11.3, Page 372 Degrees of Freedom

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13.07776.31412.70631.82163.657 21.88562.92004.30276.96459.9250............ 101.37221.81252.22812.76383.1693............ 1001.29011.66041.98402.36422.6259 1.2821.64491.96002.32632.5758 0.80 0.90 0.950.980.99 90% confidence interval; df = n-1 = 10

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0 1.8125 P(t > 1.8125) =.05 -1.8125.05.90 t 10 P(t < -1.8125) =.05

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Conf. leveln = 30 z = 1.64590%t = 1.6991 z = 1.9695%t = 2.0452 z = 2.3398%t = 2.4620 z = 2.5899%t = 2.7564

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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 Peterson’s rushing yards in each game: 84 60 86 102 88 79 153 123 182 171 108 210 154 212 86 199 Construct a 95% confidence interval for Peterson’s mean rushing ABILITY during the 2012-2013 season.

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

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Determining Sample Size to Estimate

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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.

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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?

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Confidence Interval for

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Good news: we have an equation Bad news: 1. 1.Need to know s 2. 2.We don’t know n so we don’t know the degrees of freedom to find t * n-1

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A Way Around this Problem: Use the Standard Normal

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.95 Confidence level Sampling distribution of x

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Estimating s Previously collected data or prior knowledge of the population If the population is normal or near- normal, then s can be conservatively estimated by s range 6 99.7% of obs. Within 3 of the mean

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Example: sample size to estimate mean height µ of NCSU undergrad. male students We want to be 95% confident that we are within.5 inch of so ME =.5; z*=1.96 Suppose previous data indicates that s is about 2 inches. n= [(1.96)(2)/(.5)] 2 = 61.47 We should sample 62 male students

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Example: Sample Size to Estimate a Population Mean -Textbooks Suppose 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.

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Example: Sample Size to Estimate a Population Mean -NFL footballs The manufacturer of NFL footballs uses a machine to inflate new footballs The 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?

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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 psi So range =13.7 – 13.3 =.4; range/6 =.4/6 =.067 12348...

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Chapter 23 Hypothesis Tests for a Population Mean 33

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25 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?

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As in any hypothesis tests, a hypothesis test for requires a few steps: 1.State the null and alternative hypotheses (H 0 versus H A ) a)Decide on a one-sided or two-sided test 2.Calculate the test statistic t and determine its degrees of freedom 3.Find the area under the t distribution with the t-table or technology 4.Determine the P-value with technology (or find bounds on the P-value) and interpret the result

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Step 1: State the null and alternative hypotheses (H 0 versus H A ) Decide on a one-sided or two-sided test H 0 : = versus H A : > (1 –tail test) H 0 : = versus H A : < (1 –tail test) H 0 : = versus H A : ≠ –tail test) Step 2:

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We perform a hypothesis test with null hypothesis H 0 : = 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.

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38 The one-sample t-test; P-Values Recall: The P-value is the probability, calculated assuming the null hypothesis H 0 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 is H A : 0 ) or 2-tailed (that is, the alternative hypothesis is H A : ≠ 0 ).

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39 P-Values If H A : > 0, then P-value=P(t > t 0 ) Assume the value of the test statistic t is t 0 If H A : < 0, then P-value=P(t < t 0 ) If H A : ≠ 0, then P-value=2P(t > |t 0 |)

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25 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? H 0 : μ = 95.92 H A : μ 95.92 where is the average fastball velocity of the top 25 2013 pitchers

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n = 25; df = 24 Reject H 0 : Since P-value <.05, there is sufficient evidence that top 25 pitchers in 2013 on average throw harder 0.008 2. 58 H 0 : μ = 95.92 H A : μ 95.92 t, 24 df Conf. Level0.10.30.50.70.80.90.950.980.99 Two Tail0.90.70.50.30.20.10.050.020.01 One Tail0.450.350.250.150.10.050.0250.010.005 dfValues of t 240.12700.39000.68481.05931.31781.71092.06392.49222.7969 P-value =.008

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42 2.4922 < t = 2.58 < 2.7969; thus 0.01 < p < 0.005. 0.008 2. 58 t, 24 df

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A 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%? H 0 : μ = 10 H A : μ 10 where μ is true unknown mean percentage of unpopped kernels

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n = 8; df = 7 Reject H 0 : there is sufficient evidence that true mean percentage of unpopped kernels is less than 10%.02 0 H 0 : μ = 10 H A : μ 10 -2. 51 t, 7 df Exact P-value =.02 Conf. Level0.10.30.50.70.80.90.950.980.99 Two Tail0.90.70.50.30.20.10.050.020.01 One Tail0.450.350.250.150.10.050.0250.010.005 dfValues of t 70.13030.40150.71111.11921.41491.89462.36462.99803.4995 2.3646 < |t| = 2.51 < 2.9980 so.01 < P-value <.025

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