Probability Models, Confidence Intervals - Introduction to Statistics | MATH 216, Exams of Mathematics

Download Probability Models, Confidence Intervals - Introduction to Statistics | MATH 216 and more Exams Mathematics in PDF only on Docsity! Probability Models and Confidence Intervals Math 216 – Introduction to Statistics Discrete Probability Models I. Binomial Probability Model Used when we want to know how many trials are necessary until we achieve the desired outcome and are given a sample size that is less than or equal to 30 Variables p = probability of success q = probability of failure (1-p) x = number of trials until the first success occurs n = sample size Formulas Expected value (mean):  = np Standard Deviation:  = npq Probability of a success after exactly x outcomes: P(x) = xx qp x n       Example 1 A national study shows that approximately 6% of all people have type O-Negative blood. In a study of 16 people, find: a. The expected number of the 16 people who have O-Negative blood b. The standard deviation of this sample c. The probability that none of the 16 have O-negative blood d. The probability that 1 or two of the 16 have O-Negative blood Example 2 A statewide study shows that 22% of all people have been to a minor-league baseball game this year. In a study of 21 people, find: a. The expected number of the 21 people who have been to a minor league game b. The standard deviation of this sample c. The probability that exactly 6 of the 21 have been to a minor league game d. The probability that 3 or 4 of the 21 have been to a minor league game II. Approximating the Binomial Distribution using the Normal Model Used when we want to know how many trials are necessary until we achieve the desired outcome and are given a sample size that is greater than 30 Continuity Correction Factor - Must add or subtract 0.5 to retain the original value in the probability distribution. Variables p = probability of success q = probability of failure (1-p) x = number of trials until the first success occurs n = sample size Formulas Expected value (mean):  = np Standard Deviation:  = npq Using z =  x to find a z-score, we can convert this to an area under the normal curve Example 1 A national study shows that approximately 6% of all people have type O-Negative blood. In a study of 170 people, find: a. The expected number of the 170 people who have O-Negative blood b. The standard deviation of this sample c. The probability that less than 9 of the170 have O-negative blood d. The probability that between 8 and 12 the 170 have O-Negative blood Example 2 A statewide study shows that 22% of all people have been to a minor-league baseball game this year. In a study of 250 people, find: a. The expected number of the 250 who have been to a minor league game b. The standard deviation of this sample c. The probability that exactly 60 of the 250 have been to a minor league game d. The probability that more than 70 of the 250 have been to a minor league game III. Poisson Probability Distribution - Implemented when we only know the average number of discrete occurrences in a given interval P(x) = ! xe x   where   mean number of occurrences in given interval x = desired number of outcomes Examples A confidence interval is a range of values computed uses a sample. The confidence interval is an estimate where the true population mean or proportion falls within with a specific level of certainty. Confidence Interval for Means: Population Standard Deviation is Known - CI for mean uses the following critical values for *z based upon the confidence level: 90%  1.645 95%  1.96 99%  2.575 - Population standard deviation must be known Steps for computing a CI for one mean when population standard deviation is known: Find *z based upon confidence level ME = n z   * Compute the CI by n zxCI   * Example A sample of 225 students showed an average GPA of 2.56. The population standard deviation of all students is known to be 0.39. - Compute the margin of error at 99% confidence - Create a 99% CI for the true GPA of all students. Explain what it means. - Create a 90% CI for the true GPA of all students. Explain what it means. Confidence Interval for One Proportion 90%  1.645 95%  1.96 99%  2.575 Steps for computing CI of one proportion Find *z based upon confidence level Margin of Error = ME = n pq z * pn szpCI  *% Example A recent study performed by HCC showed that of 300 randomly selected beginning algebra students, 67 did not pass. Based on this data, answer the following: a. Compute the margin of error at 90% confidence b. Construct a 90% CI for the true proportion of student who fail beginning algebra. Explain what the CI means c. Construct a 99% CI for the true proportion of student who fail beginning algebra, explain what it means and compare it to 90%CI Confidence Intervals for Means: Population Standard Deviation is Unknown - CI for mean where population standard deviation unknown uses the T-Distribution to generate the critical value *( )t - The mean and standard deviation of the sample must be given - *t depends upon the confidence level and the degrees of freedom (df) - df = n - 1 Steps for computing a CI for one mean when population standard deviation is unknown: 1) Find *t from Student-T table 2) ME = xst  * where n s s x  3) Compute the CI by xstxCI  * Example 1 A sample of 13 people showed an average sleep time of 7.3 hours. While the population standard deviation is unknown, the sample standard deviation was found to be 1.9 hours. - Compute the margin of error at 95% confidence. - Create a 95% CI for the true mean sleep time of all people. Explain what it means. - Create a 99% CI for the true mean sleep time of all people. Explain what it means. Example 2 A sample of 21 cats showed an average weight of 12.31 lbs. While the population standard deviation is unknown, the sample standard deviation was found to be 3 lbs. - Compute the margin of error at 90% confidence. - Create a 90% CI for the true weight of cats. Explain what it means. - Create a 99% CI for the true weight of cats. Explain what it means. Hypothesis Testing A method of determining if an established population mean or proportion holds true with respect to a collection sample data The null hypothesis is the what is true or what is believed to be true about a population The alternative hypothesis is what is suggested to be true by sample data Hypotheses Testing of One Proportion Formulas & Variables p - Population proportion p – Sample proportion p = n pq p pp z     P-Value Conditions: 1) np ≥ 10, nq ≥ 10 2) Sample is random Alpha (significance) levels - Alpha levels tell us the cutoff point for accepting a null hypothesis - If the we are performing a two-tailed test, we have to divide the alpha level by two to encompass both tails of the normal distribution - A low alpha level will accept more p-values (i.e. more null hypotheses) Fail to reject and rejecting the null hypothesis - When the p-value is in the reject region, we reject the null hypothesis and accept the alternative. - When the p-value is in the fail to reject region we fail to reject the null hypothesis and reject the alternative hypothesis Steps for One Proportion Hypothesis testing 1) Determine the null and Alternative Hypotheses 2) Check the conditions to use the test 3) Find the p-value using standard z-score methods 4) Fail to reject or reject the null hypothesis based on the p-value and the alpha level. State your conclusion and answer what has been asked. Example 1 A nationwide study of grocery store goods stated that about 5% of all food products were outdated. At the Weis market in Abingdon, 11 of 515 randomly selected food items were deemed outdated. Is there sufficient evidence to suggest that fewer items are outdated at the Weis Market in Abingdon than the national average? Let us test at the 5% significance level. Example 2 In 2000, a study of HCC algebra students says that 65% of them pass on their first attempt. In 2004, a survey of 450 algebra students reported that 287 passed on their first attempt. Is there sufficient evidence to suggest that the number of students that passed in 2004 is different than in 2000? Let us test at the 10% significance level. x – Sample mean s – Sample SD n – Sample size df – Degrees of freedom (n-1)  – Significance level t* – Critical value (from table) Formula for test value n s x t   Conditions: 1) Random – Samples must be randomly chosen 2) Independent – Samples must be independent 3) Nearly Normal – Data must follow a nearly normal distribution Steps for One Mean Hypothesis testing when Population Standard Deviation is Unknown 1) Determine the null and alternative hypotheses 2) Check to ensure conditions hold 3) Calculate x and xs by calculator. Also find df. 4) Find t* (critical value) and t (test value) 5) Fail to reject or reject the null hypothesis based on the t-value and t* Example 1 In 1998, the average price for bananas was 51 cents per pound. In 2003, the following 15 sample prices (in cents) were obtained from local markets: 50 53 55 53 50 57 58 54 48 47 50 57 57 51 55 Is there significant evidence to suggest that the average retail price of bananas is different than 51 cents per pound? Test at the 5% significance level. Example 2 A 1995 study of 4-year college graduates showed that the average graduate had 132 credit hours. The following 12 samples of credit hours were recorded from Maryland college students concerning their credit hours upon graduation: 137 135 151 122 131 141 155 132 123 130 145 130 Is there significant evidence to suggest that the average number of credit hours for 4-year graduates is more than 132 credit hours? Test at the 1% significance level. Hypothesis Test of Two Means I. Small, Independent Samples with unknown but equal population deviations Conditions 1) Population standard deviations are unknown but assumed to be equal 2) The samples are randomly selected independent 3) Two populations are approximately normally distributed or the samples are large Formulas Pooled Standard Deviation: 2 ))(1())(1( 21 2 22 2 11    nn snsn s p Estimator of Standard Deviation: 21 11 21 nn sS pxx  Test Value: 21 21 xx S xx t    Degrees of freedom (df) = 221  nn Example A car sales manager records that in June 25 cars were sold with an average price of $16,500 and a standard deviation of $2,000. In December 20 cars were sold with an average price of $17,300 and a standard deviation of $1,600. The two populations are normally distributed and the standard deviations are unknown but equal. Is there significant evidence to suggest that the average price of cars in June is less than the average price of cars in December? Test at the 5% significance level. II. Small, Independent Samples with unknown and unequal population deviations Conditions 1) Population standard deviations are unknown and unequal 2) 301 n , 302 n , and are independent 3) Two populations are approximately normally distributed Formulas Degrees of Freedom: 1 )( 1 )( )()( 2 2 2 2 2 1 2 1 2 1 2 2 2 2 1 2 1                              n n s n n s n s n s df  rounded down! Estimator of Standard Deviation: 2 2 2 1 2 1 )()( 21 n s n s S xx   Test Value: 21 21 xx S xx t    Example The Highway Administration shows that a study of 14 random cars on I-95 went an average of 73 mph with a standard deviation of 5 mph. A study of 20 randomly selected cars on I-83 showed an average speed of 69.5 mph with a standard deviation of 4 mph. The two populations are normally distributed and the standard deviations are unknown but unequal. Is there significant evidence to suggest that the average speed on the two highways is different? Test at the 1% significance level. III. Paired Samples Conditions 1) n < 30 2) Population standard deviation is unknown 3) Two populations are approximately normally distributed Variables d mean of the paired differences ds sample standard deviation of the paired differences n = number of paired samples *t critical value Formulas   1 2 2      n n d d s d valuetest n s xx t d    21 Example A company claims that its 12-week program significantly reduces weight. The following 4 samples were taken showing the before/after weights of 4 randomly selected customers: Before 180 195 177 221 After 183 187 161 204 Is there significant evidence to suggest that the weights of the customers is significantly less than when they began the 12-week program? Test at the 1% significance level. The Chi-Square Distribution Chi Square Distribution Formulas df = k – 1 where k is the number of categories E EO 22 )(  O = observed value E = np = Expected Value   n x n T n T n T SSB 2 3 2 3 2 2 2 1 2 1 ... )()()(           SSW =          ... )()()( 3 2 3 2 2 2 1 2 12 n T n T n T x MSB = 1k SSB and MSW = kn SSW  MSW MSB F  where DF of numerator = k – 1 and DF of denominator = n – k x = a single observation k = # of different samples n = Total samples iT Sum of observations in sample i in number of observations in sample i Example 1 Three brands of cigarettes were tested to see if there was a difference in their tar contents (in mg). The results were as follows: Cancer-Stix Cardio Stop Tar-Max 14 10 11 16 11 15 12 22 19 18 19 18 11 9 18 1T 2T 3T  x Is there evidence to suggest that there is a significant difference in the average content of tar for the three brands? Test at the 5% significance level. Non-Parametric Statistics - Nonparametric Statistics is the branch of statistics that that does not require any information about the population to perform a test upon sample data. Signed Test for Paired Samples - Used most often to determine if there is an increase or decrease after some treatment has been applied. Formulas p = population proportion (p = 0.5 unless stated otherwise) n pq p  ValueP px z p     where signstotalofnumber signdesiredofnumber x  Steps for Sign Test for Paired Samples 1) Determine the null and alternative hypothesis 2) Compute sign table for matched pairs 3) Compute p and p-value 4) Derive the conclusion Example 1 An after school tutoring program wishes to determine if it is helping students with its program in Algebra. A random sample was taken in which 12 students take an entry test and after 2 weeks of tutoring the 12 students take a test covering the same concepts. The following are the results: Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Before 55 60 30 61 73 62 After 61 67 0 80 73 80 Student 7 Student 8 Student 9 Student 10 Student 11 Student 12 Before 60 51 0 70 53 80 After 45 70 56 70 85 100 Is there significant evidence to suggest that the after school program is beneficial? Test at the 5% significance level. Example 2 An national educational supervisor wishes to determine if the high school dropout rate is higher among males or females. The following data represents 15 regions across the U.S. and their corresponding dropout rate per 100 students: Region 1 Region 1 Region 1 Region 1 Region 1 Region 1 Male 7.3 7.5 7.7 21.8 4.2 12.2 Female 7.5 6.4 6.0 20.0 2.6 5.2 Region 1 Region 1 Region 1 Region 1 Region 1 Region 1 Male 3.5 4.2 8.0 9.7 14.1 3.6 Female 3.1 4.9 12.1 10.8 15.6 6.3 Region 1 Region 1 Region 1 Male 3.6 4.0 5.2 Female 4.0 3.9 9.8 Is there significant evidence to suggest that the dropout rates are different? Test at the 10% significance level. Spearman Rank Correlation - Used to determine if two different groups rank the quality of observations in the same manner. Tests for one of two properties: - Monotone Increasing: As one variable increases, so does the other - Monotone Decreasing: As one variable decreases so does the other Formulas )1( 6 1 2 2    nn d rs Steps for Spearman Rank Correlation 1) Derive the null and alternative hypothesis 2) Create a rank table and compute 2d 3) Compute sr and find the critical value from the table 4) Derive and state the conclusion Example 1 An army psychologist gives 2 tests to 7 randomly selected cadets: one an aggression test the other a humor test. The following is a table of the results: Cadet 1 Cadet 2 Cadet 3 Cadet 4 Cadet 5 Cadet 6 Cadet 7 Aggression 60 85 78 90 93 45 51 Humor 78 42 68 53 62 50 76