Section 9.5 Day 2

Wed and Thurs 1st period: meet in Room D174 Wed 5th period meet in Room D174

Key Question ?

Do we really have two independent samples Key Question Do we really have two independent samples or

Do we really have two independent samples Key Question Do we really have two independent samples or do we have only one sample of paired data?

Components of Test What are the components of a significance test of the mean difference based on paired observations?

Components What are the components of a significance test of the mean difference based on paired observations? 1) Name test and check conditions

Components What are the components of a significance test of the mean difference based on paired observations? 1) Name test and check conditions 2) State hypotheses

Components What are the components of a significance test of the mean difference based on paired observations? 1) Name test and check conditions 2) State hypotheses 3) Compute test statistic, find P-value, and draw sketch

Components What are the components of a significance test of the mean difference based on paired observations? 1) Name test and check conditions 2) State hypotheses 3) Compute test statistic, find P-value, and draw sketch 4) Write conclusion linked to computations and in context of problem

Name Test Name: One-sided significance test of a mean difference based on paired observations or Two-sided significance test of a mean difference based on paired observations

Check Conditions Same conditions as for a confidence interval

Check Conditions 1) Randomness: For survey, must have a random sample from one population where a “unit” may consist of a pair of twins or same person’s two feet for example. For experiment, treatments randomly assigned within each pair. If same subject assigned both treatments, treatments must be assigned in random order.

Check Conditions 2) Normality: The differences, not the two original samples, must look like it’s reasonable to assume they came from normally distributed population or sample size must be large enough that sampling distribution of sample mean difference is approximately normal. 15/40 guideline can be applied to the differences.

Check Conditions 3) Sample size: For survey, size of the population of differences should be at least ten times as large as the sample size. Condition does not apply to experiment.

State Hypotheses

State Hypotheses Null hypothesis almost always is that in the entire population, the mean of the differences is 0.

State Hypotheses Null hypothesis almost always is that in the entire population, the mean of the differences is 0. In symbols, Ho: = 0

State Hypotheses Alternative hypothesis is one of these: Ha: 0

Compute test statistic, find P-value, and draw sketch P-value is based on n - 1 degrees of freedom, where n is the number of differences.

Compute test statistic, find P-value, and draw sketch Use T-Test.

Write conclusion linked to computations and in context of problem Reject null hypothesis if P-value is less than the significance level,

Write conclusion linked to computations and in context of problem Reject null hypothesis if P-value is less than the significance level, Do not reject null hypothesis if P-value is greater than or equal to the significance level,

Write conclusion linked to computations and in context of problem Reject null hypothesis if P-value is less than the significance level, Do not reject null hypothesis if P-value is greater than or equal to the significance level, Write a conclusion that relates to specific situation

Page 652, P33

Page 652, P33 The repeated measures design shows the strongest relationship, as it should, because measurements are made on the same person.

Page 652, P33 The repeated measures design shows the strongest relationship, as it should, because measurements are made on the same person. The scatterplot for the completely randomized design shows the weakest relationship, or very little association between the measurements.

Page 652, P33(b) One-sided significance test for the difference of two means

Page 652, P33(b) Conditions: Two treatments were randomly assigned to the available experimental units.

Page 652, P33(b) Two treatments were randomly assigned to the available experimental units. Boxplots fairly symmetric with no outliers so reasonable to assume both samples came from normally distributed populations. Note: if not given in problem, then you must show your graphs

Page 652, P33(b) Two treatments were randomly assigned to the available experimental units. Boxplots fairly symmetric with no outliers so reasonable to assume data came from normally distributed populations. Experiment so only two conditions to check.

Ho:

Ho: µstand = µsit , where µstand is the true mean pulse rate for standing and µsit is the true mean pulse rate for sitting for this group of subjects Ha:

Ho: µstand = µsit , where µstand is the true mean pulse rate for standing and µsit is the true mean pulse rate for sitting for this group of subjects Ha: µstand > µsit standing increases mean pulse rate

Completely randomized design Page 652, P33(b) Completely randomized design 2-SampTTest Inpt: Data Stats x1: 75.71 sx1: 11.68 Pooled: No Yes n1: 14 Calculate x2: 64.57 sx2: 9.33 n2: 14

Completely randomized design Page 652, P33(b) Completely randomized design t 2.788; P-value 0.005

Page 652, P33(b) I reject the null hypothesis because the P-value of 0.005 is less than the significance level of 0.05.

Page 652, P33(b) I reject the null hypothesis because the P-value of 0.005 is less than the significance level of 0.05. If all the students could have been given each of the standing and sitting treatments, there is sufficient evidence to support the claim that standing increases the mean pulse rate compared to sitting.

Matched Pairs Design (MPD) Page 652, P33(c) Matched Pairs Design (MPD)

Name Test Name: One-sided significance test of a mean difference based on paired observations

Page 652, P33(c) Matched Pairs Design Check conditions: Treatments randomly assigned within each matched pair.

Page 652, P33(c) Matched Pairs Design Check conditions: Treatments randomly assigned within each matched pair. The boxplot of the differences shows skewness, but not enough to be a serious problem for a sample this large. There are no outliers. So it’s reasonable to assume the differences came from a normally distributed population

Page 652, P33(c) Matched Pairs Design Ho: µd = 0, where µd is the mean difference in pulse rates if each treatment could have been assigned to each subject within each pair

Page 652, P33(c) Matched Pairs Design Ho: µd = 0, where µd is the mean difference in pulse rates if each treatment could have been assigned to each subject within each pair Ha: µd > 0

Page 652, P33(c) Matched Pairs Design T-Test Inpt: Data Stats : 0 List: L3 Freq: 1 Calculate

Page 652, P33(c) Matched Pairs Design t 3.421; P-value 0.0023

Page 652, P33(c) Matched Pairs Design I reject the null hypothesis because the P-value of 0.0023 is less than the significance level of 0.05.

Page 652, P33(c) If each treatment could have been assigned Matched Pairs Design I reject the null hypothesis because the P-value of 0.0023 is less than the significance level of 0.05. If each treatment could have been assigned to each subject in each pair, there is sufficient evidence to support the claim that standing increases the mean pulse rate compared to sitting.

Repeated Measures Design (RMD) Page 652, P33 (d) Repeated Measures Design (RMD)

Name Test Name: One-sided significance test of a mean difference based on paired observations

Repeated Measures Design (RMD) Page 652, P33 (d) Repeated Measures Design (RMD) Check conditions: Both treatments randomly assigned to each subject

Repeated Measures Design (RMD) Page 652, P33 (d) Repeated Measures Design (RMD) Check conditions: Both treatments randomly assigned to each subject The boxplot of the differences shows little skewness so it’s reasonable to assume the differences came from a normally distributed population.

Page 652, P33 (d) Repeated Measures Design (RMD) Ho: µd = 0, where µd is the mean difference in pulse rates if each treatment could have been given in both orders to each subject Ha: µd > 0

Repeated Measures Design Page 652, P33(d) Repeated Measures Design T-Test Inpt: Data Stats : 0 List: L3 Freq: 1 Calculate

Repeated Measures Design (RMD) Page 652, P33 (d) Repeated Measures Design (RMD) t 6.80; P-value = 1.324 X 10 - 7 The test statistic is quite large and would be difficult to see in a sketch.

Repeated Measures Design (RMD) Page 652, P33 (d) Repeated Measures Design (RMD) I reject the null hypothesis because the P-value of 1.324 X 10 - 7 is less than the significance level of 0.05.

Repeated Measures Design (RMD) Page 652, P33 (d) Repeated Measures Design (RMD) I reject the null hypothesis because the P-value of 1.324 X 10 - 7 is less than the significance level of 0.05. There is sufficient evidence to support the claim that standing increases the mean pulse rate compared to sitting.

Page 652, P33 (e) In this study, all three designs produced statistically significant results, with decreasing P-values of 0.005, 0.0023, and less than 0.0001. So, the evidence in favor of the alternative hypothesis increased as you reduced variation between pairs by first matching on base pulse rate and then using repeated measures. This is fairly typical of the performance of these three designs.

Page 653, P34 (a)

Page 653, P34 (a) There is no reason to pair a particular bird with a particular fish, so these are independent samples.

Page 655, E70

Page 655, E70 a. You do not have two independent samples; because the measurements of the distance are for the fifth launch and tenth launch for each team, this is a repeated measures design. You should use a one-sample test of the mean of the differences.

Page 655, E70 a. You do not have two independent samples; because the measurements of the distance are for the fifth launch and tenth launch for each team, this is a repeated measures design. You should use a one-sample test of the mean of the differences. A one-sided test should be used because you want to test whether the teams improved, which would mean the distance on the tenth launch would be longer.

Page 655, E70 b. Check conditions: This is not a random sample of teams and the paired observations are not in random order; this is an observational study of teams.

Page 655, E70 b. Check conditions: This is not a random sample of teams and the paired observations are not in random order; this is an observational study of teams. The boxplot of the differences is basically symmetric so can assume differences came from normally distributed population.

Page 655, E70 b. Check conditions: We do not know if there were 60 or more teams that launched gummy bears.

Page 655, E70 b. The conditions have not been met for a significance test for paired differences but the test result can still be useful.

Page 655, E70 State your hypotheses: H0 : μd = 0, where μd is the mean of the differences between the distance on the tenth launch and the distance on the fifth launch Ha :

Page 655, E70 State your hypotheses: H0 : μd = 0, where μd is the mean of the differences between the distance on the tenth launch and the distance on the fifth launch Ha : μd > 0

Page 655, E70

Page 655, E70 Write your conclusion in context, linked to your computations: I do not reject the null hypothesis because the P-value of 0.0998 is greater than = 0.01. There is not statistically significant evidence that the teams improved from launch 5 to launch 10.

Page 655, E70 c. What type of error might we have made?

Page 655, E70

Page 655, E70 d. The result may not match your intuition. You would still expect some improvement with practice, although you would expect less improvement from the fifth launch to the tenth than from the first launch to the tenth. At some point you might expect that more practice is not going to cause any more noticeable improvement, but you might not expect that to occur already after the fifth launch.

Questions?