AP Statistics Chapter 11 Section 1

TestConfidence Interval FormulasAssumptions 1-sample z-test mean SRS Normal pop. Or large n (n>40) Know 1-sample t-test mean (could also be used for a matched pairs design) SRS < very normal 15 < n < can be used except in the presence of outliers or strong skewness > use regardless of normality 2-sample z-test 2-sample t-test 1-proportion TESTConfidence Interval Formula(s)Assumptions 2-proportion Goodness of fit 1-way table

Unrealistic Assumption, known Assumptions we make in order to do inference about a population mean: 1.SRS 2.Observations from the population have a normal distribution with mean and standard deviation. Both are unknown parameters.

S Because we do not know the true population standard deviation we estimate with the sample standard deviation, s. This is called the standard error instead of the standard deviation.

T-distribution When the standard error is used a new distribution is formed. It is not normal. It is called the t- distribution.

T versus Z T says the same thing z says: How far from the mean of the population the mean of the sample falls in standard error/deviation units. There is a different t-distribution for each sample size – therefore, each one has a specific degree of freedom. T-distributions are symmetric about zero and are bell shaped As the degrees of freedom increase the t-distribution becomes more and more like the normal distribution. Pic on page 589 Table C gives values for t-scores.

Cokes lose sweetness? 2.0, 0.4, 0.7, 2.0, -0.4, 2.2, -1.3, 1.2, 1.1, 2.3 Is this good evidence that the cokes lost sweetness? Given that the cokes did not lose sweetness the observed loss of 1.02 would occur less than 2 out of every 100 times just by chance therefore, this statistically significant evidence rejects that the cokes did not lose sweetness.

TeacherPre-TestPost-TestGain

Did the summer training significantly improve the teachers’ comprehension of spoken French? Given that the summer training had no impact on the teachers’ comprehension of spoken French the observed difference of 2.5 would occur approx. 5 out of every 10,000 times just by chance therefore this evidence rejects that the summer training had no impact on the teachers’ comprehension of spoken French. 95% Confidence Interval (1.146, 3.854)

What is the probability that a type 1 error was made in the last problem? Type 1 error – reject the null hypothesis when it is true This error says that the summer institute did make a difference for these teachers when in fact it did not. Did not list an alpha level but reported a 95% confidence interval – There is a 5% chance that this test will reject that the summer institute has no impact on the skills of hs French teachers when it really does not have an impact on their skills.

What is the probability of a type II error in this problem? Type II error – accept the null hypothesis when it is not true Accept that the summer institute does not make a difference in the skills of HS French teachers when it does. Power The probability that this test will accept the lack of the summer institute’s effectiveness when it really is effective is approx. ( ) or.1056.

What does power of.8944 mean? This test has an approx. 89% ability to detect any deviations from the lack of effectiveness of this summer institute that are present.