 # Chapter 18: Inference about One Population Mean STAT 1450.

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Chapter 18: Inference about One Population Mean STAT 1450

Connecting Chapter 18 to our Current Knowledge of Statistics ▸ We know the basics of confidence interval estimation (Chapter 14) and tests of significance (Chapter 15). Nuances that we should be aware of were also presented (Chapter 16). 18.0 Inference about One Population Mean

Parameters and their Point Estimates ▸ In the coming chapters, we will either find confidence intervals for the population parameters, or, conduct tests of significance regarding their hypothesized values. ▸ In either case, the point estimates will help us in our endeavors. 18.0 Inference about One Population Mean Measure Sample Statistic and Point Estimate Population Parameter Mean μ Standard Deviation sσ Proportion of Successes

Inference when σ is unknown 18.0 Inference about One Population Mean

Inference when σ is unknown 18.0 Inference about One Population Mean

Conditions for Inference about a Mean ▸ The conditions for inference about a mean are listed on page 437 of the text. Random sample:  Do we have a random sample?  If not, is the sample representative of the population?  If not a representative sample, was it a randomized experiment? 18.1 Conditions for Inference about a Population Mean

Conditions for Inference about a Mean Large enough population : sample ratio:  Is the population of interest ≥ 20 times ‘n’? The population is from a Normal Distribution.  If the population is not from a Normal Distribution, then the sample size must be “large enough” with a shape similar to the Normal Distribution; then we apply the Central Limit Theorem. 18.1 Conditions for Inference about a Population Mean

Standard Error 18.1 Conditions for Inference about a Population Mean

Example: Standard Error ▸ A random sample of 49 students reported receiving an average of 7.2 hours of sleep nightly with a standard deviation of 1.74. What is the standard error of the mean? 18.1 Conditions for Inference about a Population Mean

Example: Standard Error 18.1 Conditions for Inference about a Population Mean

The t-distribution 18.2 The t Distributions

Standard Error ▸ Now the sample standard deviation will replace s; allowing us to use the one-sample t statistic for confidence intervals and tests of significance. ▸ As mentioned earlier, the t-distribution is “not quite Normal.” 18.1 Conditions for Inference about a Population Mean

The t Distributions ▸ Here is a plot of two t distributions (dashed) and the standard Normal distribution (solid): 18.2 The t Distributions

T-distribution Compared to the Z-Distribution 18.2 The t Distributions SimilaritiesDifferences Symmetric about 0T has thicker tails Single-peaked & Bell- shaped Varies based upon ‘degrees of freedom’

Using Table C ▸ 18.2 The t Distributions

Using Table C Notes: 1. The t-distribution critical values decrease as the degrees of freedom (df) increase. 2. The final row includes 1.645, 1.96, & 2.576. These are “common confidence levels” & z*. 3. Confidence intervals based upon “t” will be slightly wider than those based upon “z.” 4. Be conservative. When the exact df is not listed, “round down” and use the closest df that does not exceed the df that is desired. 18.2 The t Distributions

The One-sample t Confidence Interval 18.3 The One-sample t Confidence Interval

The One-sample t Confidence Interval 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels ▸ Hemoglobin (Hb) levels are normally distributed, and should neither be too large, nor too small. A random sample of 11 boys from an underserved country had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Compute a 90% confidence interval for the average hemoglobin level for boys from this particular country. 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components Do we have an SRS? Yes. Stated as a random sample. Steps for Success- Constructing Confidence Intervals for  (  unknown). 1.Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components Do we have an SRS? Yes. Stated as a random sample. Large enough population: sample ratio? Yes.. Steps for Success- Constructing Confidence Intervals for  (  unknown). 1.Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components Do we have an SRS? Yes. Stated as a random sample. Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. Steps for Success- Constructing Confidence Intervals for  (  unknown). 1.Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components Do we have an SRS? Yes. Stated as a random sample. Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. t-distribution? Steps for Success- Constructing Confidence Intervals for  (  unknown). 1.Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components Do we have an SRS? Yes. Stated as a random sample. Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. t-distribution? Yes. n = 11 < 40 … Steps for Success- Constructing Confidence Intervals for  (  unknown). 1.Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 1. Components Do we have an SRS? Yes. Stated as a random sample. Large enough population: sample ratio? Yes. N > 20*n=20*11=220 N=Population of boys > 220. t-distribution? Yes. n = 11 < 40 … but data is approximately Normal, so we can use t-distribution. Steps for Success- Constructing Confidence Intervals for  (  unknown). 1.Confirm that the 3 key conditions are satisfied (SRS?, N:n?, t-distribution?). 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 90% df=11-1=10 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Example: 90% CI for Hb levels 18.3 One-Sample t Confidence Intervals

Technology Tips – Computing Confidence Intervals (  unknown) 18.3 One-Sample t Confidence Intervals

Larger n, narrower interval.

The One-sample t Test of Significance 18.4 One-Sample t Test

The One-sample t Test of Significance 18.4 One-Sample t Test

The One-sample t Test of Significance 18.4 One-Sample t Test

Example: Hemoglobin levels ▸ Recall from earlier that Hb levels are normally distributed. Our original example featured a random sample of 11 boys from an underserved country that had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Is there significant evidence, at the.05 level of significance that the average Hb level for boys from this country is below 12, which results in ___________? 18.4 One-Sample t Test

Example: Hemoglobin levels ▸ Recall from earlier that Hb levels are normally distributed. Our original example featured a random sample of 11 boys from an underserved country that had an average hemoglobin level of 11.3 g/dl with a standard deviation of 1.5. Is there significant evidence, at the.05 level of significance that the average Hb level for boys from this country is below 12, which results in ___anemia________? 18.4 One-Sample t Test

Steps for Success – Conducting Tests of Significance 15.4 Tests for a Population Mean Steps for Success- Conducting Tests of Significance 1.Set up your Hypotheses. 2.Check your Conditions. 3. Compute the Test Statistic. 4. Compute the P-Value. 5. Make a Decision.

Example: Hemoglobin levels 18.4 One-Sample t Test State: Are boys from this underserved country anemic (i.e., Hb  < 12 g/dl)?

Example: Hemoglobin levels 18.4 One-Sample t Test State: Are boys from this underserved country anemic (i.e., Hb  < 12 g/dl)? Plan: a.) Identify the parameter. µ= mean Hb level for boys from this underserved country.

Example: Hemoglobin levels 18.4 One-Sample t Test

Example: Hemoglobin levels 18.4 One-Sample t Test

Example: Hemoglobin levels 18.4 One-Sample t Test

Example: Hemoglobin levels 18.4 One-Sample t Test

Example: Hemoglobin levels 18.4 One-Sample t Test

Example: Hemoglobin levels 18.4 One-Sample t Test

Example: Hemoglobin levels 18.4 One-Sample t Test Plan: f) Sketch the region(s) of “extremely unlikely” test statistics.

Example: Hemoglobin levels 18.4 One-Sample t Test Solve:

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: a)Check the conditions for the test you plan to use.  Random Sample? Yes. Stated as a random sample.

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: a)Check the conditions for the test you plan to use.  Random Sample? Yes. Stated as a random sample.  Large enough population: sample ratio? Yes. The number of boys is arbitrarily large; therefore, N > 20*11 = 220.

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: a)Check the conditions for the test you plan to use.  Random Sample? Yes. Stated as a random sample.  Large enough population: sample ratio? Yes. The number of boys is arbitrarily large; therefore, N > 20*11 = 220.  Large enough sample; Normal or t-distribution? Yes. n = 11 < 40. But data is approximately Normal, so we can use t-distribution.

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic c)Determine (or approximate) the P-Value. 1.548  -1.372 > -1.548 > -1.812 P-value

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic c)Determine (or approximate) the P-Value. 1.548  -1.372 > -1.548 > -1.812 .10 > P-value >.05 P-value

Example: Hemoglobin levels 18.4 One-Sample t Test Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

Example: Hemoglobin levels 18.4 One-Sample t Test Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

Example: Hemoglobin levels 18.4 One-Sample t Test Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate P-value is larger than 0.05, we fail to reject the null hypothesis.

Example: Hemoglobin levels 18.4 One-Sample t Test Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate P-value is larger than 0.05, we fail to reject the null hypothesis. b) Interpret the decision in the context of the original claim.

Example: Hemoglobin levels 18.4 One-Sample t Test Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate P-value is larger than 0.05, we fail to reject the null hypothesis. b) Interpret the decision in the context of the original claim. There is not enough evidence (at  =.05) that, boys from this country are typically anemic (i.e., mean Hb <12 g/dl).

Technology Tips – Conducting Tests of Significance (  unknown) 18.4 One-Sample t Test

Example: Hemoglobin levels (n= 11, 20, and 33) ▸ Let’s now use technology to conduct the test of significance at  =.05 for the three different sample sizes (n =11, 20, and 33). We will particularly focus on the test statistic, p-value, and decision at  =.05. 18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance (σ unknown) 18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance (σ unknown) 18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance (σ unknown) 18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance (σ unknown) 18.4 One-Sample t Test

Technology Tips – Conducting Tests of Significance (σ unknown) 18.4 One-Sample t Test

Matched Pairs t Procedures 18.6 Matched Pairs t Procedures ▸ One way to demonstrate that a treatment causes an observed effect is to use a matched pairs experiment. ▸ In a matched pairs design  subjects are matched in pairs and each treatment is given to one subject in each pair or  observations are taken on the same subject before-and-after some treatment.

Matched Pairs t Procedures 18.6 Matched Pairs t Procedures ▸ To compare the responses to the two treatments in a matched pairs design, find the difference between the responses within each pair. Then apply the one-sample t procedures to these differences.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Some researchers claim that music relaxes students and reduces stress while studying. 12 students were selected at random. Their initial resting pulse rate (beats/minute) was obtained, and each person participated in a month-long music-listening, relaxation therapy program. A final resting pulse rate was taken at the end of the experiment. The data are given below. ▸ Is there any evidence that music reduced the mean pulse rate, and consequently, reduced stress? Assume the underlying distributions are normal and use a 0.025 level of significance. SubjectAmyBobCalDeeEveFayGusHalIkeJoeKelMoe Initial Pulse Rate 677167837075716872887870 Final Pulse Rate 617270765861745961647177

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:  D= “Diff” < 0 (implies that the pulse rate has “reduced.”)

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:  D= “Diff” < 0 (implies that the pulse rate has “reduced.”)  Final Pulse Rate – Initial Pulse Rate < 0

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:  D= “Diff” < 0 (implies that the pulse rate has “reduced.”)  Final Pulse Rate – Initial Pulse Rate < 0  Final Pulse Rate < Initial Pulse Rate

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Music “reduces” pulse rate implies: Change in Pulse rate = Difference in Pulse Rates = Final Pulse Rate – Initial Pulse Rate All three of these expressions are equivalent.:  D= “Diff” < 0 (implies that the pulse rate has “reduced.”)  Final Pulse Rate – Initial Pulse Rate < 0  Final Pulse Rate < Initial Pulse Rate  Diff Final-Initial = Diff Final<Initial this is preferred since it directly aligns with the hypothesis presented.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures SubjectAmyBobCalDeeEveFayGusHalIkeJoeKelMoe Initial Pulse Rate 677167837075716872887870 Final Pulse Rate 617270765861745961647177 D (final-initial) -613-7-12-143-9-11-24-77 ▸ Compute the sample statistics for the Diff Final-Initial check the distribution for our assumptions.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ State: What is the practical question that requires a statistical test? Does music reduce stress (as measured by pulse rates)? µ Final - Initial < 0 or µ D < 0 where D = Final Pulse Rate – Initial Pulse Rate

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Plan: a.Identify the parameter.  = mean difference between final and initial pulse rates

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures

Example: Music relaxation therapy 18.4 One-Sample t Test Solve: a)Check the conditions for the test you plan to use.  Random Sample? Yes. Stated as a random sample.

Example: Music relaxation therapy 18.4 One-Sample t Test Solve: a)Check the conditions for the test you plan to use.  Random Sample? Yes. Stated as a random sample.  Large enough population: sample ratio? Yes. The number of students is arbitrarily large; therefore, N > 20*12 = 240.

Example: Music relaxation therapy 18.4 One-Sample t Test Solve: a)Check the conditions for the test you plan to use.  Random Sample? Yes. Stated as a random sample.  Large enough population: sample ratio? Yes. The number of students is arbitrarily large; therefore, N > 20*12 = 240.  Large enough sample; Normal or t-distribution? Yes. n = 12 < 40. But the data seem to be approximately Normal, so we will attempt to use the t-distribution.

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic c)Determine (or approximate) the P-Value.

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic c)Determine (or approximate) the P-Value. 2.511  -2.328 > -2.511 >-2.718

Example: Hemoglobin levels 18.4 One-Sample t Test Solve: b)Calculate the test statistic c)Determine (or approximate) the P-Value. 2.511  -2.328 > -2.511 >-2.718 .02 > P-Value >.01 P-value

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ).

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate P-value is less than 0.025, we reject the null hypothesis.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate P-value is less than 0.025, we reject the null hypothesis. b) Interpret the decision in the context of the original claim.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Conclude: a) Make a decision about the null hypothesis (Reject H 0 or Fail to reject H 0 ). Because the approximate P-value is less than 0.025, we reject the null hypothesis. b) Interpret the decision in the context of the original claim. There is enough evidence at the  =.025 level to conclude that music reduces stress (and lowers pulse rates).

Constructing a 95% confidence interval for the average difference in initial and final pulse rates. 18.6 Matched Pairs t Procedures

Constructing a 95% confidence interval for the average difference in initial and final pulse rates. 18.6 Matched Pairs t Procedures

Constructing a 95% confidence interval for the average difference in initial and final pulse rates. 18.6 Matched Pairs t Procedures

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11,.95)=2.201 by looking across the df = 11 row and down the 95% confidence level column.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11,.95)=2.201 by looking across the df = 11 row and down the 95% confidence level column.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11,.95)=2.201 by looking across the df = 11 row and down the 95% confidence level column.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Construct a 95% confidence interval for the average difference in initial and final pulse rates. Because we rejected the null hypothesis, we will now construct a 95% confidence interval for the difference in pulse rates. We find t*(11,.95)=2.201 by looking across the df = 11 row and down the 95% confidence level column. We are 95% confident that the mean difference in final and initial pulse rates (with Final minus Initial) is between -0.785 and -11.881 bpm. Or, we are 95% confident that participating in the relaxation experiment reduce one’s pulse rate by an average of 0.7852 to 11.881 bpm.

Example: Music relaxation therapy 18.6 Matched Pairs t Procedures ▸ Technology output for the hypothesis test and confidence interval:

Robustness of t Procedures ▸ A confidence interval or significance test is called robust if the confidence interval or P-value does not change very much when the conditions for use of the procedure are violated. ▸ Except in the case of small samples, the condition that the data are an SRS from the population of interest is more important than the condition that the population distribution is Normal. 18.7 Robustness of t Procedures

Robustness of t Procedures ▸ The t-procedures guard against non-Normality except when there is strong skewness or outliers present. ▸ When the data are not from a Normal distribution we also need to consider the sample size: 18.7 Robustness of t Procedures Sample size less than 15The t procedures can be used if the data close to Normal (roughly symmetric, single peak, no outliers)? If there is clear skewness or outliers then, do not use t. Sample size between 15 and 40 The t procedures can be used except in the presences of outliers or strong skewness. Sample size is at least 40The t procedures can be used even for clearly skewed distributions.

Robustness of t Procedures ▸ Note that we have changed the “large enough sample” condition to be adaptable to the situations that we encounter. This is because t procedures are robust against violations of Normality. 18.7 Robustness of t Procedures

Robustness of t Procedures ▸ Example: The number of text messages sent daily for 25 college students are below. Can we safely use t-procedures? 18.7 Robustness of t Procedures 10182530103 10222542118 12232651120 15242753130 17252875135

Robustness of t Procedures ▸ Example: The number of text messages sent daily for 25 college students are below. Can we safely use t-procedures? ▸ No. The sample size is 25, but there are clear outliers and strong skewness. 18.7 Robustness of t Procedures 10182530103 10222542118 12232651120 15242753130 17252875135

Five-Minute Summary ▸ List at least 3 concepts that had the most impact on your knowledge of inference about a population mean. _____________________________________________

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