A new drug designed for treatment of hormone secretion problems is tested on laboratory rats with hormone secretion problems.The scientists are concerned that although many of the secretion problems were solved using the drug, there is an undesirableadditional outcome of the weight of the laboratory rats increasing after the new drug is tested on them. A random sample of fourrats that were treated with the drug was weighed before and after treatment where the weight is given in grams.RatBefore treatmentAfter treatment145044023520720600 7204600630Assuming that the change in weight, (after treatment - before treatment) for the population of all rats treated with the newdrug is Normally distributed with mean μ, a 95% confidence interval of μ is25.0 39.60.O 25.0 56.09.25.0+ 64.31.Sedang

Before After450440520600720720600630

Answered: A new drug designed for treatment of…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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2. The city council is considering a law that would ban concealed weapons in all public facilities. A sample has been selected from the community and surveyed about support for the proposed ordinance. Is there a statistically significant relationship between age and support for the law against carrying concealed weapons (use p=.05, χ2(critical) = 3.841)? Use the five step model as a guide and write a sentence or two interpreting your results. Age Under 40 40 and older Support For 142 77 219 Against 106 170 276 248 247 495
10% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea. Of the 369 people from the inner city surveyed, 22 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.01? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: Ho: ? μ p Select an answer = < ≠ > (please enter a decimal) H1: ? μ p Select an answer > ≠ < = (Please enter a decimal) The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject accept fail to reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population proportion is not significantly smaller than 10% at αα = 0.01, so…
We consider the problem of improving lithium car batteries. A research group is makıng great advances using a new type anode and they claim that the mean life is greater than 1600 recharge cycles. To support this claim, they create 36 new batteries and subject them to recharge cycles until they fail. The claim will be established if the sample mean lifetime is greater than 1640 cycles. Otherwise, the claim will not be established and further improvements are needed. Assume that the population standard deviation o = 192 cycles is known. (a) Find the probability of a Type I error. (b) Find the probability of a Type II error when u = 1680 cycles.
If you have a research scenario in which the IV has three or more levels and is between subjects in nature with a quantitative DV, then you would use the One Way Between Subjects ANOVA. O True O False
14% of all Americans suffer from sleep apnea. A researcher suspects that a lower percentage of those who live in the inner city have sleep apnea. Of the 359 people from the inner city surveyed, 43 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.05? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: Ho: ? μ p Select an answer = ≠ > < (please enter a decimal) H1: ? p μ Select an answer < > = ≠ (Please enter a decimal) The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject fail to reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest the populaton proportion is significantly smaller than 14% at αα = 0.05, so…
You have determined that in the past three years there 7 of the 14 were hand injuries. Previous projections had indicated that under normal conditions, only four hand injuries were expected and 12 other injuries were expected. The remaining injuries were due to other causes. Calculate the chi-square value and discuss whether there was a significant difference in observed data vs. the expected data. Also discuss how the process of hypothesis testing might prove helpful o = 7 hand injuries and 7 other injuriese = 4 expected hand injuries and 12 other injuries
A study is being conducted to compare vitamin C and zinc to determine which is better at fighting colds. Customers believe vitamin C is better at fighting colds. What are the appropriate hypotheses for this testing scenario? Let μc equal the mean of the effectiveness of vitamin C and μz equal the mean of the effectiveness of zinc. O Ho: Hc - Hz= 0 Ha Hc - Hz > 0 O Ho: Hc - Hz= 0 Ha Mc-Hz 0 o Ho: Xe – xz = 0 Hai Xi — Xz #0
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 0.002
Fil 10. A small pilot study is conducted to investigate the effect of a nutritional supplement on total body weight. Six participants agree to take the nutritional supplement. To assess its effect on body weight, weights are measured before starting the supplementation and then after 6 weeks. The data are shown below. Is there a significant increase in body weight following supplementation? Run the test at a 5% level of significance. Demonstrate the five steps for hypothesis testing and explain your results. Subject İnitial Weight Weight after 6 Weeks 1 155 157 2 142 145 3 176 180 4 180 175 210 209 6. 125 126
When only two treatments are involved, ANOVA and the Student’s t-test (Chapter 11) result in the same conclusions. Also, for computed test statistics, t2 = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of six students who took the course in the normal lecture format. The other group of eight students took the course in a distance format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups. Traditional Lecture Distance 36 45 31 35 40 45 34 35 33 45 38 36 43 41 a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p-value, F crit to 4 decimal places.) Source of Variation SS df MS F p-value F crit Treatment 96.01 1 96.01 5.67 0.0350 Error 203.21 12 16.93 Total 299.21 13…

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