(a) Are the lives of these brands of batteries different? Conduct a hypothesis test. (b) Analyze the residuals from this experiment.
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Three brands of batteries are under study. It is suspected that the lives (in weeks) of the three brands are different. Five randomly selected batteries of each brand are tested with the following results:
Weeks of Life |
||
Brand 1 |
Brand 2 |
Brand 3 |
100 |
76 |
108 |
96 |
80 |
100 |
92 |
75 |
96 |
96 |
84 |
98 |
92 |
82 |
100 |
(a) Are the lives of these brands of batteries different? Conduct a hypothesis test.
(b) Analyze the residuals from this experiment.
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- In a population-based cohort study, an entire community was interviewed regarding smoking habits and then followed for one year. Upon ascertainment of all lung cancer deaths, the investigator obtained the following data: Number of Individuals Lung Cancer Deaths Smokers 24,500 15 Nonsmokers 10,500 2 Calculate the risk difference per 100,000 per year. Round to the tenth decimaA random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 20 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 20 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained. Treated with multivitamin Treated with vitamin C Days to recover from flu 3.2, 7.3, 8.4, 6.1, 2.5, 6.7, 6.3, 3.7, 7.0, 7.7, 9.1, 9.9, 9.0, 2.6, 2.0, 5.1, 5.7, 2.5, 2.9, 4.5 2.7, 5.3, 3.7, 5.5, 4.5, 3.8, 7.6, 5.3, 3.6, 5.2, 2.3, 4.7, 3.6, 5.2, 2.3, 7.1, 3.3, 6.1, 7.9, 4.2 Send data to calculator. Send data to Excel Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the population standard…A college alumni Organization sent a survey to all recent graduates to ask their annual income. 20% of the alumni responded, and their mean annual income was $40,000 with a standard deviation of $10,000. Explain why these data should not be used to construct a confidence interval for the mean annual income of all recent graduates.
- A two-way ANOVA experiment with interaction was conducted. Factor A had three levels (columns), factor B had five levels (rows), and six observations were obtained for each combination. Assume normality in the underlying populations. The results include the following sum of squares terms: SST=1542 SSA= 1042 SSB = 358 SSAB = 39 a. Construct an ANOVA table. (Round "MS" to 4 decimal places and "F" to 3 decimal places.) ANOVA Source of Variation Rows Columns Interaction Error Total SS 0 df Answer is not complete. 0 MS F p- value 0.000 0.000 0.002Consider a simple logistic model in which the response variable is attendance at a conference (yes = 1; no = 0), and the sole independent variable is a college degree (degree = 1; no degree = 0). The estimated odds ratio for the college degree variable is 1.25. What is its proper interpretation?A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 20 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 20 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained. Days to recover from flu Treated with 8.5, 5.6, 8.8. 6.1. 6.1. 3.3. 8.9, 6.0, 3.3, 9.7, 6.0, 9.3, 2.3, 4.6. 5.5.6.7. 6.6. 4.9. 10.0.6.6 multivitamin 6.5, 5.0, 3.2, 4.7, 3.5, 2.7, 3.8, 4.4, 4.6, 3.6, 4.0, 2.7, 3.2, 4.8, 6.3, 7.5, 6.9, 4.7, 4.8, 1.3 Treated with vitamin C Send data to calculator Send data to Excel Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the population standard…
- Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year many small plots of equal size but different soil/fertilizer conditions are planted with wheat. At the end of the growing season, the yield (in pounds) of the wheat on the plot is measured. Suppose for a random sample of years, one plot gave the following annual wheat production (in pounds): 3.81 2.97 5.19 3.40 3.05 2.40 4.35 3.00 4.89 5.23 3.37 4.00 3.84 5.42 5.09 5.03 For this plot, the sample variance is . Another random sample of years for a second plot gave the following annual wheat production (in pounds): 3.53 3.65 3.89 4.19 3.44 3.42 3.58 3.83 3.46 4.32 3.58 3.45 3.77 3.54 3.48 4.07 For this plot, the sample variance is . Test the claim using that the population variance of annual wheat production for the first plot is larger than that for the second plot. Find (or estimate) the P-value of the sample…A random selection of volunteers at a research institute have been exposed to a typical cold virus. After they started to have cold symptoms, 15 of them were given multivitamin tablets formulated to fight cold symptoms. The remaining 15 volunteers were given placebo tablets. For each individual, the length of time taken to recover from the cold is recorded. At the end of the experiment the following data are obtained. Days to recover from a cold Treated with multivitamin Treated with placebo 3.0, 5.6, 1.5, 6.8, 3.8, 7.5, 5.8, 4.6, 2.4, 5.0, 7.5, 5.0, 2.6, 1.7, 6.7 4.9, 6.1, 4.9, 4.2, 3.4, 5.5, 5.6, 3.4, 7.9, 6.8, 4.8, 4.2, 5.7, 2.2, 4.0 Send data to Excel Send data to calculator It is known that the population standard deviation of recovery time from a cold is 1.8 days when treated with multivitamin tablets, and the population standard deviation of recovery time from a cold is 1.5 days when treated with placebo tablets. It is also known that both populations are approximately normally…Question #6: A researcher recruits 8 people and randomly assigns them to take an herbal supplement (n, = 4) or a control (n2 = 4) for 30 days. At the end of the 30 days, each person takes the standardized memory test. The data is below: Herbal Supplement Control 22 24 26 20 19 21 27 25 How do you transform the sample statistic (M1 – Mo) into a test statistic (tobs)? Your answer
- USE THE 8-STEP HYPOTHESIS TESTING Refer to the study by Carter et al. [A-9], who investigated the effect of age at onset of bipolar disorder on the course of the illness. One of the variables studied was subjects' family history. Table 3.4.1 shows the frequency of a family history of mood disorders in the two groups of interest: early age at onset (18 years or younger) and later age at onset (later than 18 years). Family History of Mood Disorders Early ≤ 18(E) 28 19 41 53 Later > 18 (L) 35 38 44 60 141 Total 63 57 85 113 Negative (A) Bipolar disorder (B) Unipolar (C) Unipolar and bipolar (D) Total 177 318 Source: Tasha D. Carter, Emanuela Mundo, Sagar V. Parkh, and James L. Kennedy, "Early Age at Onset as a Risk Factor for Poor Outcome of Bipolar Disorder," Journal of Psychiatric Research, 37 (2003), 297-303. Can we conclude on the basis of these data that subjects 18 or younger differ from subjects older than 18 with respect to family histories of mood disorders? Let a = .05.A sunscreen company is attempting to improve upon their formula so that it lasts in water longer. They have 4 lead scientists who each came up with a different formulas. In order to see if there is a difference in the time the sunscreen lasts the CEO collects a random sample of each of the four sunscreens the data is shown below. Test the claim that at least one sunscreen has a different lifespan in water at a 0.10 level of significance. Sunscreen A Sunscreen B Sunscreen C Sunscreen D 84 33 31 71 75 66 31 78 57 72 43 63 64 77 64 89 43 52 46 84 76 34 61 45 The hypotheses for this ANOVA test would be: Η 0: μ Αμ Bμ c μ D HA : At least one mean is different. (claim) 0.10 Complete the ANOVA table below: (round answers to 3 decimal places) df MS F p-value Between Within The decision of the test is to: reject Ho do not reject H0