1 / 20

Lecture 17 Section 8.2

Lecture 17 Section 8.2. Objectives: Tests concerning hypotheses about means One sample tests Two sided tests and Confidence Intervals Two sample tests (independent samples) Paired t-tests. Tests for a population mean.

tyler-mejia
Download Presentation

Lecture 17 Section 8.2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 17Section 8.2 Objectives: • Tests concerning hypotheses about means • One sample tests • Two sided tests and Confidence Intervals • Two sample tests (independent samples) • Paired t-tests

  2. Tests for a population mean To test the hypothesis H0 : µ = µ0 based on an SRS of size n from a Normal population with unknown mean µ and known standard deviation σ, we rely on the properties of the sampling distribution N(µ, σ/√n). The P-value is the area under the sampling distribution for values at least as extreme, in the direction of Ha, as that of our random sample. Again, we first calculate a z-value and then use Table A. Sampling distribution σ/√n µdefined by H0

  3. P-value in one-sided and two-sided tests One-sided (one-tailed) test Two-sided (two-tailed) test To calculate the P-value for a two-sided test, use the symmetry of the normal curve. Find the P-value for a one-sided test and double it.

  4. One sample t-test Suppose that an SRS of size n is drawn from an N(µ, σ) population. When s is known, the sampling distribution of x bar is N(m, s/√n). When s is estimated from the sample standard deviation s, the sampling distribution follows at distribution t(m, s/√n) with degrees of freedom n − 1. is the one-sample t statistic.

  5. The P-value is the probability, if H0 is true, of randomly drawing a sample like the one obtained or more extreme, in the direction of Ha. The P-value is calculated as the corresponding area under the curve, one-tailed or two-tailed depending on Ha: One-sided (one-tailed) Two-sided (two-tailed)

  6. Example The medical director of one large company is concerned about the effects of stress on the company’s younger executives. The mean systolic blood pressure for males 35 to 44 years of age (national average) is 128 and standard deviation is 15. They examine the records of 72 executives in this age group and finds that their mean systolic blood pressure is 129.93. Assume that the population distribution is normal. Is this evidence that the mean blood pressure for all the company’s young male executives is higher than the national average? Test this at α=0.05 significance level. • State hypotheses. b. Compute test statistic and P-value. c. Make a decision and state a conclusion in terms of the problem.

  7. Example The recommended daily dietary (RDA) allowance for zinc among males older than 50 years is 15mg/day. The following data on zinc intake for a sample of males age 65-74 years: n=115, sample mean=11.3, sample sd=6.43 Does this suggest that true average daily zinc intake for the entire population of males age 65-74 is less than the recommended allowance? Use α=0.05.

  8. Example Consider the following data on the fill amounts in "16oz." ketchup bottles: 15.39, 15.62, 16.05, 15.90, 15.47, 15.83, 15.80, 15.65. Assume that the population distribution of the fill amount is normal with unknown σ . Test H0: μ = 15.5 vs. Ha: μ > 15.5 at the α = 0.01 significance level.

  9. Confidence intervals to test hypotheses Because a two-sided test is symmetrical, you can also use a confidence interval to test a two-sided hypothesis. Packs of cherry tomatoes (σ= 5 g): H0 : µ= 227 g versus Ha : µ ≠ 227 g Sample average 222 g. 95% CI for µ = 222 ± 1.96*5/√4 = 222 g ± 4.9 g 227 g does not belong to the 95% CI (217.1 to 226.9 g). Thus, we reject H0. In a two-sided test, C = 1 – α. C confidence level α significance level α /2 α /2

  10. Ex: Your sample gives a 99% confidence interval of . With 99% confidence, could samples be from populations with µ = 0.86? µ = 0.85? Logic of confidence interval test Cannot reject H0: m = 0.85 Reject H0 : m = 0.86 99% C.I. A confidence interval gives a black and white answer: Reject or don't reject H0. But it also estimates a range of likely values for the true population mean µ. A P-value quantifies how strong the evidence is against the H0. But if you reject H0, it doesn’t provide any information about the true population mean µ.

  11. Example Consider the following weights of some runners are expressed in kilograms. 67.8 61.9 63.0 53.1 62.3 59.7 55.4 58.9 60.9 69.2 63.7 68.3 64.7 65.6 56.0 57.8 66.0 62.9 53.6 65.0 55.8 60.4 69.3 61.7 Assume that the population (weights of all runners) has normal distribution with mean μ (unknown) and the population standard deviation σ = 4.5 kg. • Give a 95% confidence interval for the mean weight of the population of all such runners. b. Based on this confidence interval, does a test of H0: μ = 61.3 kg Ha: μ ≠ 61.3 kg reject H0 at the 5% significance level?

  12. Comparing two samples (B) Population Sample 2 Sample 1 (A) Population 1 Population 2 Sample 2 Sample 1 Which is it? We often compare two treatments used on independent samples. Is the difference between both treatments due only to variations from the random sampling (B), or does it reflect a true difference in population means (A)? Independent samples: Subjects in one samples are completely unrelated to subjects in the other sample.

  13. Two-sample z statistic We have two independent SRSs (simple random samples) possibly coming from two distinct populations with (m1,s1) and (m2,s2). We use 1 and 2 to estimate the unknown m1 and m2. When both populations are normal, the sampling distribution of ( 1- 2) is also normal, with standard deviation : Then the two-sample z statistic has the standard normal N(0, 1) sampling distribution.

  14. Two independent samples t distribution We have two independent SRSs (simple random samples) possibly coming from two distinct populations with (m1,s1) and (m2,s2) unknown. We use ( 1,s1) and ( 2,s2) to estimate (m1,s1) and (m2,s2), respectively. To compare the means, both populations should be normally distributed. However, in practice, it is enough that the two distributions have similar shapes and that the sample data contain no strong outliers.

  15. Two-sample t significance test The null hypothesis is that both population means m1 and m2 are equal, thus their difference is equal to zero. H0: m1 = m2 <=> m1 − m2 = 0 with either a one-sided or a two-sided alternative hypothesis. We find how many standard errors (SE) away from (m1 − m2) is ( 1− 2) by standardizing with t: Because in a two-sample test H0poses (m1 − m2) = 0, we simply use

  16. Example Consider the lifetimes of two kinds of light bulbs: • Take a random sample of size n1=40 from the population with mean μ1 and standard deviation σ1=26, • Take independently another random sample of size n2=50 from the population with mean μ2 and standard deviation σ2=22. Test : H0: μ1 −μ2 =0 vs. Ha: μ1 −μ2 ≠ 0 at α = .05 .

  17. Example In the comparison of two kinds of paint, a consumer testing service finds that 34 1-gallon cans of Benjamin Moore paint cover on the average 5480 square feet with a standard deviation of 62 feet, whereas 41 1-gallon cans of Pittsburgh paint cover on the average 5452 square feet with a standard deviation of 51 feet. Test to see whether or not the Pittsburgh paints cover a larger area on average, at the 1% significance level.

  18. Example The following data on tensile strength (psi) of linear specimens both when a certain fusion process was used and when this process was not used: 1. No fusion: 2784 2700 2655 2822 2511 3149 3257 3213 3220 2753 2. Fused: 3027 3356 3359 3297 3125 2910 2889 2902 Test if the fusion process increased the average tensile strength at the 5% significance level.

  19. In these cases, we use the paired data to test the difference in the two population means. The variable studied becomes Xdifference = (X1 − X2), andH0: µdifference= 0 ; Ha: µdifference>0 (or <0, or ≠0) Conceptually, this is not different from tests on one population. Paired t-test

  20. Example The following data was obtained from a sample of n=16 subjects. Each observation is the amount of time, expressed as a proportion of total time observed, during which arm elevation was below 30o. The two measurements from each subject were obtained 18 months apart. During this period, work conditions were changed, and subjects were allowed to engage in a wider variety of work tasks • Does the data suggest that true average time during which elevation is below 30o differs after the change from what it was before the change? • Does it appear the change in work conditions decreases true average time by more than 5?

More Related