Hypothesis Testing and Confidence Intervals for Population Proportions and Means, Exams of Data Analysis & Statistical Methods

Download Hypothesis Testing and Confidence Intervals for Population Proportions and Means and more Exams Data Analysis & Statistical Methods in PDF only on Docsity! Stat 245 Recitation 19 (11/27/2007) TA: Dongmei Li 1 STAT245 Recitation Name: Summary for Chapter 10: Hypothesis Tests for a Population Proportion Procedure: Step 1: Find population characteristic of interest: and hypothesized value Step 2: Write down the null hypothesis and alternative hypothesis. H0: = hypothesized value Ha: > hypothesized value or Ha: < hypothesized value or Ha: hypothesized value Step 3: Find sample size n and sample proportion p. Step 4: Find significance level . Step 5: Check assumptions. p is the sample proportion from a random sample Both hypothesized value 10n and 1 hypothesized value 10n Step 6: If the assumptions in step 3 are satisfied, then calculate the test statistic. hypothesized value hypothesized value 1 hypothesized value z n p Step 7: Calculate P-value: The P-value is the area under the standard normal curve more extreme than the test statistic, in the direction indicated by the alternative hypothesis. Ha: > hypothesized value Area under z curve to right of calculated z Ha: < hypothesized value Area under z curve to left of calculated z Ha: hypothesized value 2(area to right of z) if z is positive, or 2(area to left of z) if z is negative. Step 8: Make a decision. The test decision is to reject H0 whenever P-value < (significance level). The common used significant level is 0.05 followed by 0.01 and 0.10. Step 9: Write down your conclusion. Stat 245 Recitation 19 (11/27/2007) TA: Dongmei Li 2 If P-value , then the sample provides convincing evidence to support the claim that If P-value > , then there is not enough evidence to support that claim that Relationship between tests and intervals o Hypothesis tests and confidence intervals have different goals but are closely related. A test assesses the degree of evidence against the hypothesis that there is no difference/effect between two groups An interval estimates the magnitude of the difference/effect between two groups o When a two-sided test results in rejecting H0 at significance level , then a CI with confidence level 100(1- ) will not include the value zero. Procedures for calculating confidence intervals for : Check Assumptions: 1. The sample is selected randomly; 2. 10np and 1 10n p Formula for calculating confidence intervals for : (z* in following formula is z critical value, remember the z critical value for 95% confidence interval is 1.96, for 99% confidence interval is 2.58 and for 90% confidence interval is 1.645.) 1p p p z n Exercise: 1. A pilot s group claims that among aborted takeoffs leading to an aircraft s going off the end of the runway, more than 5% are due to bird strikes. A random sample of 740 aborted takeoffs in which the aircraft overran the runway was taken. Among those 740 cases, 50 were due to bird strikes. Test the group s claim at the 5% level of significance and calculate a 95% confidence interval for the proportion of aircraft s going off the end of the runway due to bird strikes. H0: = 0.05 (The population proportion of aircraft s going off the end of the runway due to bird strikes is 0.05). Ha: > 0.05 (The population proportion of aircraft s going off the end of the runway due to bird strikes is more than 0.05). n=740, p=50/740 = 0.0676, significant level =0.05. n 0= 740*0.05 = 37 > 10 n(1- 0) = 740(1-0.05) = 703 > 10 Stat 245 Recitation 19 (11/27/2007) TA: Dongmei Li 5 Ha: µ > hypothesized value Area to right of calculated t under t curve with df = n -1 Ha: µ < hypothesized value Area to the left of calculated t under t curve with df = n -1 Ha: µ hypothesized value 2(area to right of t) if t is positive, or 2(area to left of t) if t is negative. Step 8: Make a decision. The test decision is to reject H0 whenever P-value < (significance level). The common used significant level is 0.05 followed by 0.01 and 0.10. Step 9: Write down your conclusion. If P-value , then the sample provides convincing evidence to support the claim that If P-value > , then there is not enough evidence to support that claim that o Confidence interval for µ procedure: s x t n (Here, t* is the t critical value that you can find from the t table, do not forget the degree of freedom is sample size n-1). This procedure is valid as long as: The sample is random Either the population is normal or the sample size is large (n>30) The interpretation is the same as before: We are C% confident that the CI contains the actual value of the population mean . This confidence means that if random samples were repeatedly drawn from the population and a CI constructed from each sample, then in the long run C% of those intervals would succeed in capturing the population mean . Exercise: 3. 50 smokers were questioned about the number of hours they sleep each day. We want to test the hypothesis that the smokers need less sleep than the general public which needs an average of 7.7 hours of sleep. If the sample mean is 7.5 and the standard deviation is 0.5, what can you conclude using a significant level of 0.01? Calculate a 99% confidence interval for population mean and interpret it. H0: µ = 7.7 (smokers need same amount of sleep as the general public). Ha: µ < 7.7 (smokers need less sleep than the general public). n= 50, x = 7.5, s = 0.5 , significant level =0.01. Stat 245 Recitation 19 (11/27/2007) TA: Dongmei Li 6 Random sample and n=50 > 30 are satisfied. 83.2 50/5.0 7.75.7 / 0 ns x t P-value = P(t < -2.83) = 0.003365 degree of freedom (d.f.) = 50 -1 = 49 Since the P-value is less than 0.01, we reject the null hypothesis. There is enough evidence to conclude that smokers need less sleep than the general public. 99% confidence interval for is: )6895.7,3105.7(01895.05.7 )071.0(68.25.7 50 5.0 )68.2(5.7* n s tx We are 99% confident that (7.3105, 7.6895) contains the actual value of the mean of smoker s sleeping time . 4. A hot tub manufacturer advertises that with its heating equipment, a temperature of 100 oF can be achieved in at most 15 min. A random sample of 25 tubs is selected, and the time necessary to achieve a 100 oF temperature is determined for each tub. The sample average time and sample standard deviation are 17.5 min and 2.2 min, respectively. Assume the population distribution is approximately normal. Does this information cast doubt on the company s claim? a. Carry out a test of hypotheses using significance level 0.05. b. Calculate a 95% confidence interval for population mean and interpret it. H0: µ = 15 (a temperature of 100 oF can be achieved in 15 min). Ha: µ > 15 (a temperature of 100 oF will be achieved in more than 15 min). n= 25, x = 17.5, s = 2.2 , significant level =0.05. Random sample and the population distribution is approximately normal are satisfied. 682.5 25/2.2 155.17 / 0 ns x t P-value = P(t > 5.682) 0 degree of freedom (d.f.) = 25 -1 = 24 Stat 245 Recitation 19 (11/27/2007) TA: Dongmei Li 7 Since the P-value is less than 0.05, we reject the null hypothesis. There is enough evidence to conclude that a temperature of 100 oF can be achieved in more than 15 min. So, we will doubt on the company s claim. 95% confidence interval for is: )4064.18,5936.16(9064.05.17 )44.0(06.25.17 25 2.2 )06.2(5.17* n s tx We are 95% confident that (16.5936, 18.4064) contains the actual value of the mean time that a temperature of 100 oF can be achieved.