Finding Critical Values and Confidence Intervals. In Exercises 5–8, use the given information to find the number of degrees of freedom, the critical values
7. Platelet Counts of Women 99% confidence; n = 147, s = 65.4.
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- The data show the population (in thousands) for a recent year of a sample of cities in South Carolina. 26 26 15 29 69 21 30 29 13 26 20 38 85 19 19 23 29 25 111 47 30 49 108 30 38 Send data to Excel Part 1 of 8 The data value 29 corresponds to the 46" percentile. Part 2 of 8 The data value 38 corresponds to the 70"n percentile. Part: 2 / 8 Part 3 of 8 The data value corresponds to the 93rd percentile.arrow_forwardA random variable takes the values 3,4,5. p(X = 3) = 0.4 and E(X) = 3.8. p(X < 4/X < 5) = %3D Select one: a. O b. Od. O d. -IN -T C.arrow_forwardAspirin II: Safety Considerations Regarding the experiment in the data frame Aspirin from the abd package, the researchers wanted to know whether or not taking aspirin affects one's risk of developing cancer. Recall that they defined their parameters as follows: p1 = the proportion of ALL individuals who would develop cancer, if all of them were to take aspirin like the subjects in the Aspirin group did. p2 = the proportion of ALL individuals who would develop cancer, if all of them were to take a placebo, like the subjects in the placebo group did. They ran the code for a two-sided significance test and got the following results: ## ## ## Inferential Procedures for the Difference of Two Proportions p1-p2:## cancer grouped by treatment ## ## ## Descriptive Results:## ## yes n estimated.prop## Aspirin 1438 19934 0.07214## Placebo 1427 19942 0.07156## ## ## Inferential Results:## ## Estimate of p1-p2: 0.0005805 ## SE(p1.hat - p2.hat): 0.002586 ## ## 95%…arrow_forward
- I sent this in earlier and the response was not correct. number 15 chapter 10. part a Please use the attached table. To test H0:u=100 versus H1:NOT EQUAL TO 100, A SIMPLE RANDOM SAMPLE OF N=21 IS OBTAINED FROM A POPULATION THAT IS KNOWN TO BE NORMALLY DISTRIBUTED. A. If x with a bar over it = 104 and s= 9.5 compute the test statistic.arrow_forwardUsing the Central Limit Theorem. In Exercises 5–8, assume that females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 “Body Data” in Appendix B). a. If 1 adult female are randomly selected, find the probability that they have pulse rates with a mean between 78 beats per minute and 90 beats per minute. b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean between 78 beats per minute and 90 beats per minute. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?arrow_forwardEvery year, the students at a school are given a musical aptitude test that rates them from 0 (no musical aptitude) to 5 (high musical aptitude). This year's results were: "Aptitude Score" "Frequency" If 0 1 2 3 4 5 sum 3 6 4 3 2 4 22 0 0 6 The mode aptitude score: 2,5 there is no mode.) 8 8 20 51 ✓o The mean aptitude score: Σ(xf) Σ(5) Please find the median aptitude score by a TI-84: 2.5 2.32 (Round to 2 decimal place.) X x (Please separate your answers by, in bimodal situation. Enter DNE ifarrow_forward
- Module 5: Ihd Use the table of values and the expected frequencies to calculate the residual values and the chi square test statistic X XYZ A 34 44 22 B 19 53 32 Expected Frequencies A 25.98 47.549 26.471 B 27.02 49.451 27.529 Residual Values (Row i and column j) Chi Square Test Statistic (0.,, – E. ) E (0 – E) x =E E Round all values to 3 decimal places Y B The Chi Square Test Statistic x =arrow_forward5.3 2.4 3.5 5.2 (Reference: Crime in the United States, Federal Bureau of Investigation.) Assume that the crime rate distribution is approximately normal in both regions. i, Use a calculator to verify that x 3.51, s, = 0.81, x, = 3.87, and 0.94. i/ Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than that in New England? Use a = 0.01. %3D Medical: Hay Fever A random sample of n, = 16 communities in western Kansas gave the following information for people under 25 years of age. x,: Rate of hay fever per 1000 population for people under 25 120 128 92 123 112 93 86 06 125 95 125 122 88 L6 Arandom sample of n, = 14 regions in western Kansas gave the following information for people over 50 years old. x,: Rate of hay fever per 1000 population for people over 50 95 110 97 112 88 85 110 115 114 68 96 (Reference: National Center for Health Statistics.) i. Use a calculator to verify that x 109.50, s, - 15.41, x2 - 99.36, and sh 11.57. ii. Assume that the…arrow_forward5 c. Test a claim that the mean amount of carbon monoxide in the air in U.S. cities is less than 2.33 parts per million. It was found that the mean amount of carbon monoxide in the air for the random sample of 64 cities is 2.39 parts per million and the standard deviation is 2.12 parts per million. At α=0.10, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed. (a) Identify the claim and state H0 and Ha. What is: H0 and Ha? The claim is the hypothesis. (b) Use technology to find the critical value(s) and identify the rejection region(s). The critical value(s) is/are t0= (Use a comma to separate answers as needed. Round to two decimal places as needed.)arrow_forward
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill