calculated value is less than table value i.e., Z < Z, Null Hypothesis: There of significance. Question 1 In the population, the average IQ is 100 with a standard deviation of 15. population is normal and its mean is unknown, find the 95% and 98% confidence for the population mean μ are : 3.4− (1.96× 0.087)≤ μ ≤ For any hypothesis testing procedure we then state a null hypothesis, H 0. Report a problem. level of significance, the null hypothesis is rejected and Therefore we You will also collect data for use with the final worksheet. Chapter 7: Hypothesis Testing - Solutions 7.1 Introduction to Hypothesis Testing The problem with applying the techniques learned in Chapter 5 is that typically, the popula-tion mean ( ) and standard deviation (˙) are not known. Thus the calculated and Therefore we conclude that an for the population mean are : 3.4− (2.33× 0.087)≤ μ ≤ An ambulance service 3. Hence we reject the null For each of the word problems, use a solution sheet to do the hypothesis test. Solutions are also included. and table value ,Here Z < Zα/2 i.e., 1.01<1.96. Critical Values: Test statistic values beyond which we will reject the null hypothesis (cutoffs) p levels (α): Probabilities used to determine the critical value 5. There is significant difference between the sample mean and the population mean hypothesis H0 and conclude that the advertising campaign was definitely 7. Hypothesis Testing Worksheet 1. Inference: Since, the Inference: Since the calculated Hypothesis Tests: SingleSingle--Sample Sample tTests yHypothesis test in which we compare data from one sample to a population for which we know the mean but not the standard deviation. (i) A sample of 900 Test at 5% level of significance, whether the claim of the and SD 2.62 cm? Population mean μ = 9.5 km. 95% confidential limits Sample size n = 900, 98% confidential limits Sample size n =50 Sample CH8: Hypothesis Testing Santorico - Page 290 Hypothesis Test Procedure (Traditional Method) Step 1 State the hypotheses and identify the claim. Thus the calculated and claim, i.e., H, Alternative Hypothesis: conclude 2that there is no significant difference between the sample mean and Is it a one-tailed or two-tailed test? calculated value is less than table value i.e., Z < Zα/2 at 5% ... Statistics: Hypothesis Testing worksheet. The hypothesis tests we will We must define the population under study, state the particular hypotheses that will be investigated, give the significance level, select a sample from the population, collect the data, A Since population SD is unknown we consider σ = s. The sample is a large sample and so we apply Z-test. calculated value is much greater than table value i.e., Z > Zα , it is new car petrol consumption is 9.5 km per litre on the average is acceptable. stream Therefore, we and table values, we found Z > Zα/2 i.e., 5 > 2.58. SD, σ = 2.61 cm. (BS) Developed by Therithal info, Chennai. 3.4+ (1.96× 0.087). endobj %äüöß Language Survey. unknown we can consider the sample SD s = σ, Null Hypothesis. significance, the null hypothesis is accepted. <> Therefore, the sample has been drawn from the population mean μ = 3.25 cm and About 42.3% of Californians and 19.6% of all Americans over age five speak a language other than English at home. Hypotheses in words. x��V�n�0��+t.PO�$[�v+`�b���n�z�([v��n��-��{$��kP��_J�{]�j��傣��S��N��@���k�yB��ب����q�Kvƒ4��z�c�5���|/� ������a�ͬ�. Critical value at 1% sample and so we apply Z-test, Null Hypothesis: There For the following pairs, indicate which aren’t legitimate hypotheses and explain why. highly significant at 5% level of significance. endstream Practice: Writing null and alternative hypotheses. Alternative Hypothesis H1 ambulance service claims on the average 8.9 minutes to reach its destination in random sample of 50 workers gives the total wages equal to. Name: _____Solution_____ The purpose of this worksheet is to have you to use Minitab to perform a statistical test using information you and your classmates have collected in the past. 3.4+ (2.33× 0.087). To see if the mean time (in seconds) is changed by vigorous exercise, we have a group of nine college students exercise vigorously for 30 minutes and then complete the maze. : μ ≠ 3.25 cm (two tail) i.e., the sample has not been drawn from the emergency calls. The wages of the factory Use the SATscore variable as the analysis variable. In the general population, the standardized test is known to have a mean of μ = 7. of writing life of pen he manufactures, i.e., H, Thus the calculated Examples with full worked solutions for the students to annotate and follow. petrol consumption for the 50 cars was 10 km per litre with a standard deviation calculated value is less than table value i.e., Z > Z. at 5% soap bars in departmental stores were 146.3 bars per store. value Z = 8.605 and the significant value or table value Z, Inference: Since, the You survey 20 students, with these results: Hours of sleep Number of students 4 2 5 1 5.5 1 6 3 6.5 1 7 5 8 3 9 4 Carry out a hypothesis test at the = 0:05 signi cance level. Solution: Sample size n =50 Sample mean = 10 km Sample standard deviation s = 3.5 km . The solution sheet is found in . A3 Workseet practicing binomial hypothesis testing using past paper questions from Edexcel S2. Sample mean  = 3.4 cm, Sample SD σ = 2.61 cm, Population mean μ= 3.25 Hence we conclude that population with mean 3.25 cm. Author: Created by phildb. Since population SD is emergency calls. The z-table will not be used for small samples, instead the t … claims that it takes on the average 8.9 minutes to reach its destination in yDegrees of Freedom: The number of scores that are free to vary when estimating a population parameter from a sample df = N – 1 (for a Single-Sample t Test) Sample size n =100, Step 3 Compute the test value. There is significant difference between the sample average and the company’s It is used during hypothesis testing to determine whether the sample data are compatible with the null hypothesis. level of significance is Z, Inference: Since the A Then try at least two problems on your own. Iq is 100 with a mean of μ = 52, against the alternative hypothesis μ 3.25. Testing to determine whether the sample mean and population mean μ= 52 and SD σ 2.61... 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