The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng/ mL found fornonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n = 918 and thetest statistic is t = 56.582. Use technology to find the P-value. Based on the result, what is thefinal conclusion? Use a significance level of 0.05.Begin by using the original claim to identify the null hypothesis and the alternative hypothesis for this test.The null hypothesis (denoted by Ho) is a statement that the value of a population parameter (suchas proportion, mean, or standard deviation) is equal to some claimed value. The alternativehypothesis (denoted by H, or Ha or Ha) is a statement that the parameter has a value that somehow differsfrom the null hypothesis.

Given that Sample size n =918 Population mean =2.84 Test statistic t =56.582

Answered: he claim is that smokers have a mean…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Big babies: The National Health Statistics Reports described a study in which a sample of 310 one-year-old baby boys were weighed. Their mean weight was 24.4 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the α = 0.10 level of significance and the P-value method with the TI-84 Plus calculator. Compute the P-value. Round the answer to at least four decimal places. P-value = ____ How do I solve this problem using the TI-84 calculator?
A researcher wants to test whether the mean height for men and women are different, using a significance level of 0.05. Which is the correct hypothesis? A Ho: P1 = P2 Ha: P1 P2 D Ho: P1 = μ2 Ha: H1 H2 O C A OF OE O B B Ho: P1 = P2 Ha: P1 P2 E Ho: P1 = μ2 Ha: 12 C Ho: P1 = P2 Ha: 1: P₁ P2 F H₂:₁ = 1₂ Ha: ₁ 2
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
The void volume within a textile fabric affects comfort, flammability, and insulation properties. Permeability of a fabric refers to the accessibility of void space to the flow of a gas or liquid. An article gave summary information on air permeability (cm3/cm2/sec) for a number of different fabric types. Consider the following data on two different types of plain-weave fabric: Fabric Type Sample Size Sample Mean Sample Standard Deviation Cotton 10 51.41 0.75 Triacetate 10 132.15 3.59 Assuming that the porosity distributions for both types of fabric are normal, let's calculate a confidence interval for the difference between true average porosity for the cotton fabric and that for the acetate fabric, using a 95% confidence level. Before the appropriate t critical value can be selected, df must be determined: df = 0.5625 10 + 12.8881 10 2 (0.5625/10)2 9 + (12.8881/10)2 9 = 1.8092 0.1849…
Salma claimed that the gas price in Blue Park neighborhood (mean of $3.65) is higher than that in Green Spring (mean of $3.59). She obtained the t-value of 2.223 after calculating data from 6 gas stations in Blue Park and 5 gas stations in Green Spring. Use the chart below and determine if Salma's claim is supported! Table 10-3. Critical values of "t" significance level p= 0.05 significance level p= 0.01 TP 1 12.706 63.657 2 4.303 9.925 3 3.182 5.841 2.776 4.604 5 2.571 4.032 6 2.447 3.707 7 2.365 3.499 2.306 3.355 9 2.262 3.250 10 2.228 3.169 11 2.201 3.106 Copy the following into your text box, fill in the blanks, and answer the questions A. The degree of freedom is: B. The t-calculated value is: and the t-critical value is : C. Comparison: t-calc t-crit (answer with ()) D. Is the null hypothesis accepted or rejected? Explain how you determined that! Also, explain what it means to reject or accept the hypothesis! E. Explain if Salma's claim is supported based on your t-value…
The claim is that smokers have a mean cotinine level greater than the level of 2.84 ng / mL found for nonsmokers. (Cotinine is used as a biomarker for exposure to nicotine.) The sample size is n= 892 and the test statistic is t = 52.657. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.10. State the null and alternative hypotheses. Ho: H (Type integers or decimals. Do not round.) The test statistic is. (Round to two decimal places as needed.) The P-value is. (Round to three decimal places as needed.) Based on the P-value, there sufficient evidence at a significance level of 0.10 to the claim that smokers have a mean cotinine level greater than the level of 2.84 ng / mL found for nonsmokers.
Big babies: The National Health Statistics Reports described a study in which a sample of 335 one-year-old baby boys were weighed. Their mean weight was 25.2 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys differs from 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the o 0.01 level of significance and P-value method with the Critical Values for the Student's t Distribution Table. (a) State the appropriate null and alternate hypotheses (b) Compute the value of the test statistic. (c) State a conclusion.
Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.02, y = 100.75, r = 0.852, P-value = 0.000, and y = 5.93 + 0.94x, where x represents the IQ score of the husband. Find the best predicted value of y given that the husband has an lQ of 100? Use a significance level of 0.05. %3D Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is Critical values of the pearson correlation coefficient r (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r X = 0.05 a = 0.01 INOTE: To test Ho: p=0 n Jagainst H,: p+0, reject Ho if the absolute value of r is greater than the critical value in the table. 4 0.950 0.990 0.878 0.959 0.811 0.917 7 0.754 0.875 0.707 0.834 0.666 0.798 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444…
A safety administration conducted crash tests of child booster seats for cars. Listed below are results from those tests, with the measurements given in hic (standard head injury condition units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified requirement? 600 662 1092 545 496 541 what are the hypothesis? identify the test statistics identify the P-value state the final conclusion that addresses the original claims what do the results suggest about the child booster seats meeting the specified requirement?
Big babies: The National Health Statistics Reports described a study in which a sample of 335 one- year-old baby boys were weighed. Their mean weight was 25.2 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys differs from 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the a= 0.01 level of significance and the critical value method with the Critical Values for the Student's t Distribution Table. Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0 μ-25 H1: u#25 This hypothesis test is a two-tailed V test. Part 2 of 5 Find the critical value(s). Round the answer(s) to three decimal places. If there is more than one critical value, separate them with commas. Critical value(s): 2.576 , 2.576 Part: 2/5 Part 3 of 5 (b) Compute the value of the test statistic. Round the answer to at least three decimal places.
Please provide a clear and complete solution. Thank you very much.

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