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1c.STATISTICAL INFERENCE
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- The first test a doctor would order to determine whether a person is infected with HIV (the virus that causes AIDS) is the ELISA test. It detects antibodies and antigens for HIV. A study in Statistical Science by J. Gastwirth estimated that, if the person is actually infected with HIV, this test produces a positive result 97.7% of the time. If a person is not infected with HIV, the test result is negative 92.6% of the time. According the the US Centers for Disease Control (CDC), an estimated 1.1 million Americans out of a population of 321 million were infected with HIV in 2015. Using the information above, determine the probability that a randomly selected person whose ELISA test is positive actually is infected with HIV? a. What is the probability that a randomly selected American is infected with HIV? b. Using the answer to part (a) and the conditional probabilities of positive and negative ELISA test results, fill out the contingency table below: ELISA Test Result…Fast reactions: In a study of reaction times, the time to respond to a visual stimulus x and the time to respond to an auditory stimulus y were recorded for each of 8 subjects. Times were measured in thousandths of a second. The results are presented in the following table. Visual Auditory 161 159 176 201 203 163 191 169 197 178 206 235 241 201 188 197 Use the P-value method to test =:H0310 versus >:H1310. Can you conclude that visual response is useful in predicting auditory response? Use the =a0.05 level of significance and TI-84 Plus calculator. Compute the least-squares regression line for predicting auditory y from visual x. Round the slope and y-intercept values to at least four decimal places. find p and t valueI need help with this
- Which type of t-test is best for the following variables:(a) Test scores of Section A51 at start of Semester (b) Test scores of Section A51 at end of Semester Paired Samples t-test Independent Samples t-test None. The variables are both discrete. None. The variables are both continuous. 8A clinical microbiology laboratory compared the frequency of isolation of DNA hybridization probe to a culture method for the sexually transmitted bacterium Neisseria gonorrhaeae. The frequency of isolation from clinical specimens was historically 10%. The lab ran 100 split samples, and the results are presented in the table below. Table. Comparison of the Distribution of Negative and Positive Results for the DNA Probe and Culture Methods. Culture Results DNA Probe Results Positive (D) Negative (D-) Positive (T) 8 4 Negative (T-) 2 92 What is the sensitivity of the test? I choose c wouldn't the sensitivity been 0.02 ? or my calculations is wrong A. 0.88 or 88% B. 0.8 or 80% C. 0.02 or 2% D. 0.81 or 81%Question 1c needed
- StatisticsStatistical significance at the 0.01 level is more difficult to achieve easier to achieve ○ less costly O less informative than significance at the 0.05 level.Data on the cost (millions of dollars) and the running time (minutes) for films of a particular season are summarized in the computer output shown to the right and the plots below. Complete parts a and b Click the icon to view the plots. a) Check the assumptions and conditions for inference Select all that apply The residual plot and the scatterplot show consistent variability The scatterplot looks straight enough The residuals look random The residuals are nearly normal None of the assumptions and conditions for inference are satisfied. b) Find a 95% confidence interval for the slope and interpret it. m% confident that the cost of making longer movies (Round to two decimal places as needed. Use ascending order.) Dependent variable is: Budget(SM) squared-24.0% R s 25.12 with 20-2-18 degrees of freedom Variable Coefficient SE(Coeff) t-ratio P-value intercept -23.5882 Run Time 0.665602 0.2788 34.12 -0.69 0.4982 2.39 0.0282 at a rate of between and milion dollars per minute Plots 120 Run…
- just part C thank you!Fisher's exact p-value 3.0 puntos posibles (calificables, resultados ocultos) In the table of data below, the column labeled Y is the observed outcome and W is the indicator of treatment, i.e. W = 1 indicates that Y in the same row is the observed outcome from a treatment unit, and W = 0 indicates that Y in the same row is the observed outcome from a control unit. Y W 0.5 1 0.2 0 0.4 0 0.7 1 Compute the test statistic T for the Fisher Exact Test. T = Conduct the two-sided Fisher Exact Test on the data above and report the p-value. Fisher's p-value:Part (i) The chance of developing breast cancer is under 11% for women. Ho: M=11, Hg μ#11 Ho: M211, Hg: μ 11 Ho: p=0.11, H₂: P# 0.11 O Ho: Ho: P 20,000, Ο Ηρ: μ 5 20,000, O Ho: p = 20,000, O Ho: p ≤ 20,000, O Ho:p> 20,000, Ho: μ ε 20.000, O Ho: p≥ 20,000, Additional Materials Reading Hg μ 5 20,000 Hg μ > 20,000 Ha: p= 20,000 Hp> 20,000 He: p ≤ 20,000 Hg με 20,000 Ha: p<20,000
