3.11 Using Statistics To Make Inferences 3 Summary Review the normal distribution Z test Z test for the sample mean t test for the sample mean Wednesday, 04 November 20155:54 PM

3.22 Goals To perform and interpret a Z test. To perform and interpret tests on the sample mean. To produce a confidence interval for the population mean. Know when to employ Z and when t. Practical Perform a t test. Perform a two sample t test, in preparation for next week.

3.33 Normal Distribution Series123 μ001 σ1½1 Tables present results for the standard normal distribution (μ=0, σ=1).

3.44 Use of Tables Prob(1≤z≤∞) = 68% of the observations Prob(-∞≤z≤-1) = lie within 1 standard deviation 0.16of the mean Prob(1.96≤z≤∞) = 95% of the observations lie Prob(-∞≤z≤-1.96) = within 1.96 standard deviations of the mean Prob(2.58≤z≤∞) = 99% of the observations lie Prob(-∞≤z≤-2.58) = within 2.58 standard deviations of the mean

3.55 Use of Tables Z Prob(1≤z≤∞) = Prob(-∞≤z≤-1) = 0.16 Prob(1.96≤z≤∞) = Prob(-∞≤z≤-1.96) = Prob(2.58≤z≤∞) = Prob(-∞≤z≤-2.58) = 0.005

3.66 Testing Hypothesis Null hypothesis H0H0 assumes that there is no real effect present Alternate hypothesis H1H1 assumes that there is some effect

3.77 Z Test For a value x taken from a population with mean μ and standard deviation σ, the Z-score is

3.88 The Central Limit Theorem When taking repeated samples of size n from the same population. 3.The distribution of the sample means approximates a Normal curve. 2.The spread of the distribution of the sample means is smaller than that of the original observations. 1.The distribution of the sample means is centred around the true population mean

3.99 Central Limit Theorem If the standard deviation of the individual observations is σ then the standard error of the sample mean value is For a sample mean,, with mean μ and standard deviation the Z-score is

Example 1 mean score 100 standard deviation 16 What is the probability a score is higher than 108? Prob(x≥108) = Prob(z≥0.5) = Z

Example 2 mean score 100 standard deviation 16 The sample mean of 25 individuals is found to be 110. The null hypothesis, no real effect present, is that μ = 100. Wish to test if the mean significantly exceeds this value.

Solution 2 Prob( ≥ 100) = Prob(z≥3.125) = , beyond our basic table Z Since the p-value is less than the result is highly significant, the null hypothesis is rejected. The sample average is significantly higher.

Estimating The Population Mean Confidence interval for the population mean Sample mean nSample size σPopulation standard deviation (known) Tabulated value of the z-score that achieves a significance level of α in a two tail test Don’t forget to multiply or divide before you add or subtract This test is not available in SPSS

Estimating The Population Mean Confidence interval for the population mean Sample mean nSample size σPopulation standard deviation (known) Tabulated value of the z-score that achieves a significance level of α in a two tail test We can be 100(1-2α)% certain the population mean lies in the interval

Normal Values Conf. level Prob. α One Tail ZαZα 90% % % Z Notation commonly used to denote Z values for confidence interval is Z α where 100(1 - 2 α ) is the desired confidence level in percent.

Example 3 standard deviation 16 mean of a sample of 25 individuals is found to be 110 Require 95% confidence interval for the population mean

Solution 3 95% sure the population mean lies in the interval [103.7,116.3]

Is there a snag?

One Sample t-Test The basic test statistic is Sample mean n Sample size s Sample standard deviation t Calculated t statistic Note now s not σ

Interpreting t-values The test has ν =n-1 degrees of freedom. If t calc < t ν (α) then we cannot reject the null hypothesis that μ=m. If t calc > t ν (α) the null hypothesis is rejected, the true mean μ differs significantly at the 2α level from m. ν the Greek letter nu Critical value from tables

If The Population Standard Deviation Is Not Available? t values with ν = n – 1 degrees of freedom ( ν the Greek letter nu) Sample mean n Sample size ν Degrees of freedom, n-1 in this case s Sample standard deviation α Proportion of occasions that the true mean lies outside the range tνtν Critical value of t from tables Don’t forget to multiply or divide before you add or subtract Note in this module, typically, the sample variance is required. Divide by n-1

If The Population Standard Deviation Is Not Available? t values Sample mean n Sample size ν Degrees of freedom, n-1 in this case s Sample standard deviation α Proportion of occasions that the true mean lies outside the range tνtν Critical value of t from tables

Two Tail t To obtain confidence limits a two tail probability is employed since it refers to the proportion of values of the population mean, both above and below the sample mean.

Example 4 An experiment results in the following estimates. Obtain a 90% confidence interval for the population mean.

Example 4 Given ν p=0.05p=0.025p=0.005p= We can be 90% (α=0.05) sure that the population mean lies in this interval [68.6,74.2].

Example 5 Claimed mean is 75 seconds, the times taken for 20 volunteers are H 0 : there is no effect so μ = 75 H 1 : μ ≠ 75 (two tail test)

Solution 5 n = 20 Σx = 1428 Σx 2 = n = 20 Σx = … = 1428 Σx 2 = … =

Solution 5 n = 20 Σx = 1428 Σx 2 =

Solution 5 n = 20 Σx = 1428 Σx 2 = s = 7.34 Note in this module, typically, the sample variance is required. Divide by n-1. To practice use mean-var.xls.mean-var.xls

Solution 5 νp=0.05p=0.025p=0.005p=0.0025p= In an attempt to “estimate” p.

Conclusion 5 Since 2.093<2.193< <p-value<0.05 (note 2α since two tail) t = There is sufficient evidence to reject H 0 at the 5% level. The experiment is not consistent with a mean of 75. In fact the 95% confidence interval is [68.0,74.8] which, as expected, excludes 75. The precise p value may be found from software.

SPSS 5 Analyze > Compare Means > One Sample t Test Note insertion of test value

SPSS 5 Basic descriptive statistics for a manual test

SPSS 5 As predicted 0.01 < p-value < 0.05 The confidence interval is to that is [67.96, 74.84].

Graph? Graph > Legacy Dialogs > Error Bar

Graph? Graph > Legacy Dialogs > Error Bar

Example 6 Experimental data Test whether these data are consistent with a population mean of H 0 is that μ = 0.250

Solution 6 νp=0.05p=0.025p=0.005p=0.0025p= t 11 (0.005)=3.106 t 11 (0.0025)=3.497 In an attempt to “estimate” p.

Conclusion 6 t = Since < < < p-value < 0.01 There is sufficient evidence to reject H 0 at the 1% level. The experimental mean would not appear to be consistent with 0.250

SPSS 6 As predicted p-value < 0.01 The confidence interval is to that is [0.26, 0.30].

Read Read Howitt and Cramer pages Read Russo (e-text) pages Read Davis and Smith pages , , ,

Practical 3 This material is available from the module web page. Module Web Page

Practical 3 This material for the practical is available. Instructions for the practical Practical 3 Material for the practical Practical 3

Whoops! From testimony by Michael Gove, British Secretary of State for Education, before their Education Committee: "Q98 Chair: [I]f 'good' requires pupil performance to exceed the national average, and if all schools must be good, how is this mathematically possible? "Michael Gove: By getting better all the time. "Q99 Chair: So it is possible, is it? "Michael Gove: It is possible to get better all the time. "Q100 Chair: Were you better at literacy than numeracy, Secretary of State? "Michael Gove: I cannot remember." Oral Evidence, British House of Commons, January 31, 2012, p. 28

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