Draw the Chi Square Curve and put in the critical value (p = 0.05)
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Draw the Chi Square Curve and put in the critical value (p = 0.05)
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- The average drying time of a manufacturer’s paint is 20 minutes. Investigating the effectiveness of a modifica-tion in the chemical composition of her paint, the manu-facturer wants to test the null hypothesis μ = 20 minutes against a suitable alternative, where μ is the average dry-ing time of the modified paint. (a) What alternative hypothesis should the manufactureruse if she does not want to make the modification in thechemical composition of the paint unless it decreases thedrying time?(b) What alternative hypothesis should the manufactureruse if the new process is actually cheaper and she wants tomake the modification unless it increases the drying timeof the paint?12) The personnel director for a small firm thinks that employees may be more likely to use sick days on Monday and Friday resulting in three-day weeks. a) What is the appropriate null and alternate hypothesis to determine if sick day absences depend on day of the week? b) The personnel director obtains sick day data for the past year and summarizes the number of absences by day of the week in the table below. Use JMP software to test the hypothesis of part a) c) Using alpha = 0.05, draw a conclusion for the hypothesis test and state the conclusion in the context of the problem. Day of Week Absences Monday 126 Tuesday 74 Wednesday 91 Thursday 82 Friday 1152. A new chemotherapy drug is released to treat leukemia and researchers suspect that the drug may have fewer side effects than the most commonly used drug to treat leukemia. The two drugs have equivalent efficacy. In order to determine if a larger study should be conducted to look into the prevalence of side effects for the two drugs, set up a Mann-Whitney U test at the alpha equals .05 level and interpret its results.Number of Reported Side-EffectsOld Drug 0 1 3 3 5New Drug 0 0 1 2 4 A) We fail to reject H0, which states the two populations are equal at the alpha equals .05 level because the calculated U value of 16.5 is greater than the critical U value of 2.B) We fail to reject H0, which states the two populations are equal at the alpha equals .05 level because the calculated U value of 8.5 is greater than the critical U value of 2.C) We reject H0 in favor of H1, which states the two populations are not equal at the alpha equals .05 level because the calculated U value of 16.5 is…
- If the alpha value is 0.05 p-value for Z test I get as 1.932146e-05 in excel or R. Null hypothesis will be rejected or accepted?An article in the Journal of the Electrochemical Society (Vol. 139, No. 2, 1992, pp. 524–532) describes anexperiment to investigate the low-pressure vapor deposition of polysilicon. The experiment was carried outin a large-capacity reactor at Sematech in Austin, Texas. The reactor has several wafer positions, and fourof these positions are selected at random. The response variable is film thickness uniformity. Threereplicates of the experiment were run, and the data are as follows:a. Test the significance of these wafer position with α=0.05.b. If proven significant, perform a multiple comparison method using Fisher’s LSD.A new chemotherapy drug is released to treat leukemia and researchers suspect that the drug may have fewer side effects than the most commonly used drug to treat leukemia. The two drugs have equivalent efficacy. In order to determine if a larger study should be conducted to look into the prevalence of side effects for the two drugs, set up a Mann-Whitney U test at the alpha equals .05 level and interpret its results. Number of Reported Side-Effects Old Drug 0 1 3 3 5 New Drug 0 0 1 2 4 A) We fail to reject H0, which states the two populations are equal at the alpha equals .05 level because the calculated Uvalue of 16.5 is greater than the critical U value of 2. B) We fail to reject H0, which states the two populations are equal at the alpha equals .05 level because the calculated Uvalue of 8.5 is greater than the critical U value of 2. C) We reject H0 in favor of H1, which…
- 1. The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype. Your task is to perform a chi-square goodness of fit test on each of two F2 data sets, and make a decision, based on your statistical analyses as to which F2 data set provides greater evidence for indicating the correct mode of inheritance. In this problem, the true mode of inheritance is autosomal dominant, homozygous lethal. Evidence is measured in the following ways: the p-value is greater than 0.05, so we do not reject the null hypothesis, and the p-value is closer to 1. A few things of which to be mindful. 1. In the parental generation, the WT parent always has a homozygous genotype. 2. For the autosomal dominant mode of inheritance, the disease gene always homozygous for the disease allele in the parental generation. 3. For the autosomal dominant, homozygous lethal any person with the…A Chi square test has been conducted to assess the relationship between marital status and church attendance. The obtained Chi square is 7.45 and the critical Chi square is 9.488. What may be concluded? Group of answer choices A.reject the null hypothesis, church attendance and marital status are independent B.fail to reject the null hypothesis, church attendance and marital status are independent C.Reject the null hypothesis, church attendance and marital status are dependent D.fail to reject the null hypothesis, church attendance and marital status are dependentThe market researcher in a certain company conducted a study to assess the effectiveness of a new computerized system of filling orders. He selected a random sample of 100 customers served using the old system and 100 served using the new system. He contacts each customer to find out whether or not the order was filled satisfactorily within 2 weeks. Results were recorded in the accompanying table: System Satisfied Not Satisfied Total OLD 68 32 100 NEW 85 15 100 Total 153 47 200 Test at alpha equal to 5% the null hypothesis that the proportion of satisfied customers among those served by the new system is the same as that among those served by the old system.
- About half of the police officers in Baltimore, Maryland, have completed a special course in community policing. As Captain of your division, you want to know if the course increased calls from the minority community for assistance from your precinct. Below is the results of before and after training in proportions for calls from the minority community in your precinct. Be sure and state whether you accept or reject the null hypothesis. Alpha is 0.05 for setting your Z critical value. After Training Before training Ps1 = 0.47 Ps2 = 0.43 N1 = 157 N2 = 113A new chemotherapy drug is released to treat leukemia and researchers suspect thst the drug may have fewer side effects than the most commonly used drug to treat leukemia. The two drugs have equivalent efficacy. In order to determine if a larger study should be conducted to look into the prevalence of side effects for the two drugs, set up a Mann-Whitney U test at the alpha equals .05 level and interpret its results. Number of reported side effects Old drug 0 1 3 3 5 New drug 0 0 1 2 41) A researcher who works for a national retail chain is interested in changes in employee satisfaction during the holiday season. The table below contains employee satisfaction scores taken at two time points (August 1st and December 1st) from a sample of 12 employees. Higher scores indicate more satisfaction. Conduct a hypothesis test to determine if there is an increase or decrease in employee satisfaction from Time 1 to Time 2. Use alpha .05. Employee Number Time 1 Satisfaction Time 2 Satisfaction10012 8.5 6.510057 6.8 410089 6.5 410126 4.2 5.710023 7 610045 6 310094 7 6.510087 3 310145 4.5 3.510023 9 8.510062 8.5 410078 4 2.5 a. State the null alternative hypothesis.b. Conduct the hypothesis test.c. State your findings.d. Report an effect size measure. Please use the formula that we used in class 2) A study was conducted to see how goal-setting setting can affect satisfaction with life. The students were randomly assigned to complete a satisfaction with life measure either…