John rejected a null hypothesis in a right-tailed test for a mean at alpha = 0.025 because the critical T value was 2.000 and his calculated T value was 2.345. We can be sure that: a.) John did not commit type 1 error B.) John committed neither type 1 nor type 11 error c.) none of the presented answers is correct because none can definitely be concluded  d.) John did not commit type 11 error

3. A random sample of 25 observations taken from a population that is normally distributed produced a sample mean of 58.5 and a standard deviation of 7.5. Find the critical and observed values of t for each of the following tests of hypothesis using a = 0.01. a) H.:µ = 55 b) H.:µ = 55 Ha:µ > 55 Ha:H # 55

A consumer group plans a comparative study of the mean life of four different brands of batteries. Ten batteries of each brand will be randomly selected and the time until the energy level falls below a pre-specified level is measured. a) Which of the following is the appropriate alternative hypothesis for the null hypothesis: H0: μa = μb = μc = μd? (In this problem, μa = mean time of Brand A, μb = mean time of Brand B, etc.) Ha: none of the means are equal Ha: μa ≠ μb ≠ μc ≠ μd Ha: μa ≠ μb, μa ≠ μc, μa ≠ μd, μb ≠ μc, μb ≠ μd, μc ≠ μd Ha: at least one of the means is different b)In order to analyze the data with ANOVA, we need to satisfy the condition of randomization. How do we know that this condition has been met? -One SRS of batteries is selected from a collection of batteries of all brands. -The batteries are randomly allocated to the four brands. -Separate random samples of batteries are selected from each brand.

If your null and alternative hypothesis are:H0:μ=9H0:μ=9H1:μ<9H1:μ<9Then the test is: left tailed right tailed two tailed

Two independent samples have been selected, 7777 observations from population 1 and 5858 observations from population 2. The sample means have been calculated to be ?⎯⎯⎯1=13.4x¯1=13.4 and ?⎯⎯⎯2=11.3x¯2=11.3. From previous experience with these populations, it is known that the variances are ?21=27σ12=27 and ?22=29σ22=29. For the hypothesis test of ?0:(?1−?2)=1.8H0:(μ1−μ2)=1.8 and ??:(?1−?2)≠1.8Ha:(μ1−μ2)≠1.8 Use ?=0.04α=0.04. (a)    Compute the test statistic. ?=z= (b)    Find the approximate p-value?−?????=p−value=   The final conclustion is A. There is not sufficient evidence to reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8. B. We can reject the null hypothesis that (?1−?2)=1.8(μ1−μ2)=1.8 and accept that (?1−?2)≠1.8(μ1−μ2)≠1.8.

If the critical z of a test is ±1.96, what inferences can you make about other aspects of the test? A) The hypothesis is non-directional and the alpha level is .05 B) The hypothesis is directional and the alpha level is .05 C) The hypothesis is directional and the alpha level is .01 D) The hypothesis is non-directional and the alpha level is .01 Please explain why

A researcher investigates whether cold medication effects mental alertness. It is known that scores on a standardized test containing a variety of problem-solving tasks are normally distributed with  = 64 and  = 8. A random sample of n = 16 teenage and a sample of n = 25 adults are given the drug and then tested. On average, the teenagers scored and average of ? = 58 and the adults scored and average of M = 65.5.a. Are the data sufficient to conclude that the medication significantly reduces mental alertness in teenagers? Test with  = .01.b. Are the data sufficient to conclude that the medication significantly increases mental alertness in adults? Test with  = .01.

Bocellie rejected a null hypothesis. The reason for doing this is that the hypothesis required a right-tailed test and the t value was 2.5 whereas the t critical value was 2. Which of the following are we certain about: Both Type I and Type II error were made A Type II error was not made Neither Type I nor Type II error was made A Type I error was not made

2. Assume that AUM body temperatures are normally distributed with o = collected a sample with the sample mean 98° F among n = mean body temperature in AUM. 1.5° F. We 9 this year. Let u be the (a) Find a two-sided 95% confidence interval for u. (b) Find an upper bounded one-sided 95% confidence interval for u. (c) Find a lower bounded one-sided 95% confidence interval for u.

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