Significance Tests in practice

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Significanc e Tests in practice Chapter 12

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Significance Tests in practice. Chapter 12. 12.1 Tests about a population mean . When we don’t know the population standard deviation σ , we perform a one sample T test instead of a Z test - PowerPoint PPT Presentation

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Significance Tests in practice

Significance Tests in practiceChapter 1212.1 Tests about a population mean When we dont know the population standard deviation , we perform a one sample T test instead of a Z test We get a T calc (instead of a Z calc) and the formula is the same except we put in the standard deviation Sx of our sample as a replacement for

Interpretation and P valuessame as Z interval, except if you are using the tables, look up T star to find the critical T value you need to beat. I recommend the calculator thoughThe P value is from your calculated T to the tail. If it is a 2 tailed test, remember this gets doubled!

T test on calcGo to Stat/Tests choose 2: T-test. Enter either data or Stats (just like Z test) Example: Diversity (P. 750)An investor with a stock portfolio sued his broker b/c lack of diversification in his portfolio led to poor performance. The table gives the rates of return for the 39 months that the account was managed by the broker. Consider the 39 monthly returns as a random sample from the monthly returns the broker would generate if he managed the account forever. Are these returns compatible with a population of =.95%, The S&P 500 average?-8.361.63-2.27-2.93-2.70-2.93-9.14-2.646.82-2.35-3.586.137-15.25-8.66-1.03-9.16-1.25-1.22-10.27-5.11-.80-1.441.28-.654.3412.22-7.21-.097.345.04-7.24-2.14-1.01-1.4112.03-2.564.332.35Step 1: Hypothesis- (Where is the mean return for all possible months that the broker could manage this account)H0: = .95 HA: .95

Step 2: Conditions- Since we do not know we must use a one-sample t test. Now we check conditions:SRS- We are told this in problemNormality: Sample is large enough (39) so CLT applies Independence: We must treat the 39 monthly returns as independent observations from the population of months in which the broker could have managed the account. Step 3: Calculations- Test Statistic: (-1.10 - .95) / (5.99/39) = -2.14 P value: df are 39-1 = 38. Since we are performing a 2-tailed test, our P-value is .039 (can be done on calc, or on table- if doing table, make sure to double the area under the curve b/c of 2 tailed!)

Step 4: Interpretation: The mean monthly return on investment for this clients account was -1.1% for this period (this is our x bar). This is significantly difference at the alpha = .05 level of the S&P 500 for the same period (t = -2.14, P 0 The mean sweetness loss for the population of tasters is positive. The cola seems to be losing sweetness in storageStep 2: Conditions- Since we do not know the standard deviation of sweetness loss in the population of tasters, we must use a one-sample t test. Now we check conditions:SRS- assume the 10 tasters are randomly selectedNormality: 10 is too small a sample so we check our normal probability plotIt is slightly left skewed, but no gaps/outliers so we proceed with caution

Independence: we assume there are more then 10x10 = 100 tasters in the population

Step 3: CalculationsTest Statistic- The basic statistics areX bardiff = 1.01 and Sdiff= 1.196Tcalc= (1.02 0)/ (1.196/10) = 2.70P value: degrees of freedom are 10-1 = 9 for alpha = .05, p = .0123

Step 4: Interpretation- A P-value this low gives strong evidence against the null hypothesis. We reject H0 and conclude that the cola has lost sweetness during storage

*Remember the 3Cs! (Conclusion, connection, context)

Robustness and PowerRecall that t procedures are robust against non-normality of the population except when outliers or strong skewness are present (skewness is more serious). As the sample size increases, the CLT ensures that our distribution of sample means approximates normality. Again, calculating power is not in this course, but remember we can still commit the same type I / type II errors when rejecting (or failing to reject) the null. 12.2 Tests about a population proportionRemember, when the 3 conditions are met (SRS, Normality, and Indep.), the sampling distribution of p hat is approximately Normal with mean equal to rho and standard deviation = (pq/n)

Note: In the standard error (denominator of test statistic) we use the values of our assumed null hypothesis- this is different from the standard error of the CI where we used sample proportion values!One proportion z on calcStat- Test 1-propZtest put in:Rhox = the count (number of successes in your sample)n = number of people in your samplechoose one-tailed or 2 tailed hypothesis calculateExample (P. 767) work stressJob stress poses a major threat to the health of workers. A national survey or restaurant employees found that 75% said that work stress had a negative impact on their personal lives. A random sample of 100 employees from a large restaurant chain finds that 68 answer yes when asked Does work stress have a negative impact on your personal life?. Is this good reason to think that the proportion of all employees in this chain who would say yes differs from the national proportion of = .75? Hypothesis: we want to test a claim about , the true proportion of this chains employees who would say that work stress has a negative impact on their personal lives. H0: =.75 HA: .75

Conditions: We should use a one-proportion Z test if the conditions are metSRS: we are told in problemNormality: The expected number of yes and no responses are (100)(.75) = 75 and (100)(.25) = 25 respectively. Both are at least 10 sIndependence: We assume the chain has at least 10(100) = 1000 employeesCalculationsTest Statistic:

P value = .1052 (if you do normalcdf, remember to double your value b/c this is a 2 tailed test! so 2x.0526!)

BUT! MUCH easier to do on Calc thoughand not use NormalCDF! Try it

Interpretation There is over a 10% chance of obtaining a sample result as unusual as or even more unusual than we did (p hat = .68) when the null is true so we have insufficient evidence to suggest that the proportion of this chain restaurants employees who suffer from work stress is different from the national survey result, .75

Confidence Intervalsusing the info from that same example, if we were to calculate a 95% CI (.59, .77)by hand, using formula:

Note: the CI standard error uses the sample proportions, but in the test statistic formula we use the value assumed in the null hypothesis

Using calc is easier though: choose 1 prop z int

Interpretation: We are 95% confident that between 59% and 77% of the restaurant chains employees feel that work stress is damaging their personal lives.