Testing Means with small samples Critical Value from the T-table –Use the one-tail heading for left or right tail Make negative for left-tail –Use the two-tails heading for two tailed test –Remember degrees of freedom For Example 1.H 0 : μ ≥ 25, n = 18, Degree of confidence = 99% 2.H 0 : μ = 110, n = 25, Degree of confidence = 95% 3.H 0 : μ ≤ 1.5, n = 12, Degree of confidence = 90% 4.H 0 : μ = 74, n = 28, Degree of confidence = 99.9%

Live example/Your Turn What would you guess to be average age of South students’ cars? –Let’s do a survey!! What would you guess to be minimum number of colleges that average South student’s applied to? –Let’s take a sample!!

Worksheet Sample mean =n =s = Original claimLabel null and alternative hypothesis Opposite claim Degree of confidence Test statistics Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept?

Worksheet Sample mean =n =s = Original claimLabel null and alternative hypothesis Opposite claim Degree of confidence Test statistics Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept?

Testing Hypothesis with Small Samples We use the T (Student) table (A-3) to find the critical value –We need to know the degrees of freedom, the significance level (alpha), and the number of tails Calculator: –[STAT] –TESTS –1: T-Test… μ 0 is the benchmark. X-bar is the mean Sx is the standard deviation n is the sample size μ select the format of H 1

Homework: Test the claims 1.Claim: Student population has a mean GPA of 2.0. A sample of 24 students has a mean is 2.35 and a standard deviation is Use a 95% degree of confidence 2.Claim: An SAT prep class produces scores above A sample of 15 students has a mean is 1685 and a standard deviation is 170. Use a 99% degree of confidence 3.Claim: The average college student needs at least 5 years to get a degree. A sample of 20 students has a mean of 5.15 years and the standard deviation is Use a 90% degree of confidence 4.The following list contains the repair costs for five BMW cars used in a controlled crash test: Use this sample to text the claim that BMW’s repair costs are under $ Using a sample of 25 adults whose mean body temperature was (standard deviation = 0.56), Test the claim that the mean body temperature for the population is 98.6

Homework #1 Sample mean = 2.35n = 24s = 0.07 Original claimµ = 2.0 (H0)Label null and alternative hypothesis Opposite claimµ ≠ 2.0 (H1) Degree of confidence95% Test statisticst = 2.45 Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept? p-value = < 0.05Reject null, reject original claim

Homework #2 Sample mean = 1685n = 15s = 170 Original claimµ > 1700 (H0)Label null and alternative hypothesis Opposite claimµ ≤ 1700 (H1) Degree of confidence99% Test statisticst = Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept? p-value = > 0.01FRT null, FTR original claim

Homework #3 Sample mean = 5.15n = 20s = 1.68 Original claimµ ≥ 5(H0)Label null and alternative hypothesis Opposite claimµ < 5 (H1) Degree of confidence90% Test statisticst = Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept? p-value = > 0.1FRT null, FRT original claim

Homework #4 Sample mean = 588n = 5s = 424 Original claimµ < 1000 (H1)Label null and alternative hypothesis Opposite claimµ ≥ 1000 (H0) Degree of confidence95% Test statisticst = Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept? p-value = < 0.05Reject null, accept original claim

Homework #5 Sample mean = 98.24n = 25s = 0.56 Original claimµ = 98.6 (H0)Label null and alternative hypothesis Opposite claimµ ≠ 98.6 (H1) Degree of confidence95% Test statisticst = Critical region Two-tailed (H 0 =) Left tailed (H 0  ) Right tailed (H 0  ) Critical value Reject or accept? p-value = < 0.05Reject null, reject original claim