) The following table shows that 100 patients infected with COVID-19 for stages 1 to 4 on a particular day were reported at Hospital XYZ. AN X Frequency 1 10 2 20 3 n 4 40
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- One way to detect an association between variables is the chi square statistic. Variables are associated if obtained chi square is greater than zero less than zero greater than 10 greater than 100The proportion of people in a given community who have Covid-19 infection is 0.005. A test is available to diagnose the disease. If a person has Covid-19, the probability that the test will produce a positive signal is 0.009 . If a person does not have the Covid-19, the probability that the test will produce a positive signal is 0.01. What is the probability that the test will generate positive signal? Which model/rule will best be good for solving the above problem and why? Comment on the types of events you see in the problem and name them.Researchers wanted to test the hypothesis that living in the country is better for your lungs than living in a city. To eliminate the possible variation due to genetic dif- ferences, they located seven pairs of identical twins with one member of each twin living in the country, the other in a city. For each person, they measured the percentage of in- haled tracer particles remaining in the lungs after one hour: the higher the percentage, the less healthy the lungs. They found that for six of the seven twin pairs the one living in the country had healthier lungs. Based on the sample size and distance between the null value and the observed proportion, estimate the strength of evidence: inconclusive, weak but suggestive, moder- ately strong, strong, or overwhelming.
- Researchers followed a random sample of 2315 middle-aged men from eastern Finland for up to 30 years. They recorded how often each man went to a sauna and whether or not he suffered sudden cardiac death (SCD). The two-way table shows the data from the study. The researchers will perform a chi-square test for independence to test the hypotheses Ho: There is no association between weekly sauna frequency and suffering from sudden cardiac death in the population of middle-aged men from eastern Finland. H: There is an association between weekly sauna frequency and suffering from sudden cardiac death in the population of middle-aged men from eastern Finland. Weekly sauna frequency What are the chi-square statistic, degrees of freedom, 1 or fewer 2-3 4 or more Total and P-value? Yes 61 119 10 190 O x' = 6.032, df = 2, P-value = 0.9510 SCD No 540 1394 191 2125 O x? = 6.032, df = 6, P-value = 0.4196 Total 601 1513 201 2315 O x? = 6.032, df = 3, P-value = 0.1101 O x? = 6.032, df = 2, P-value =…A production superintendent claims that there is no difference between the employee accident rates for the day versus the evening shifts in a large manufacturing plant. The number of accidents per day is recorded for both the day and evening shifts for n = 100 days. It is found that the number of accidents per day for the evening shift XE exceeded the corresponding number of accidents on the day shift xp on 63 of the 100 days. Do these results provide sufficient evidence to indicate that more accidents tend to occur on one shift than on the other or, equivalently, that P(XE > XD) # 1/2?A major credit card company is investigating whether the distribution of the number of credit cards used by its customers has changed from last year to this year. Customers are classified as using 1 card, 2 cards, or more than 2 cards. The company conducts a chi-square goodness-of-fit test to investigate whether there is a change in the distribution of number of cards used from last year to this year. The value of the chi-square test statistic was χ2=7.82χ2=7.82 with a corresponding pp-value of 0.02. Assuming the conditions for inference were met, which of the following is the correct interpretation of this pp-value?
- A study of fox rabies in a country gave the following information about different regions and the occurrence of rabies in each region. A random sample of n1 = 16 locations in region I gave the following information about the number of cases of fox rabies near that location. x1: Region I Data 1 9 9 9 7 8 8 1 3 3 3 2 5 1 4 6 A second random sample of n2 = 15 locations in region II gave the following information about the number of cases of fox rabies near that location. x2: Region II Data 1 1 5 1 6 8 5 4 4 4 2 2 5 6 9 What is the value of the sample test statistic? (Test the difference μ1 − μ2. Do not use rounded values. Round your final answer to three decimal places.)A glass manufacturing company wanted to investigate the effect of zone 1 lower temperature (630 vs. 650) and zone 3 upper temperature (695 vs. 715) on the roller imprint of glass. Zone 3 Upper Zone 1 Lower Roller Imprint695 630 46695 630 25695 630 47695 630 119695 650 23695 650 26695 650 47695 650 22715 630 96715 630 5715 630 26715 630 81715 650 69715 650 23715 650 5715 650 124 a. At the .05 level of significance, is there an interaction between zone 1 lower and zone 3 upper? Determine the test statistic. Fstat= Determine the p-value. p-value= b. At the .05 level of significance, is there an effect due to zone 1 lower? First, determine the hypotheses. Determine the test statistic. Fstat= Determine the p-value. p-value= c. At the .05 level of significance, is there an effect due to zone 3 upper? First, determine the hypotheses. Determine the test statistic. Fstat= Determine…
