Statistics: One Tailed vs. Two tailed?
I am doing Hypothesis testing (how many hooks are in a package of blah blah hooks?) My claim and null are M = 150 (M being mu) and my alternative is M does not equal 150. sample size = 15 P > alpha (so I accepted the null) BUT! I still do not quite understand how to determine weather or not the claim is supported, and is it is one tailed or two tailed. I need to Know: one tailed or two tailed? What value do I use to determine if the claim is correct or false. What table do I use. I know i did not describe this very well, but if you can help, please do! I need this project to go well. Finals week :( Thank you!!
College Statistics question - hypothesis testing?
ANSWER: Conclusion: H1 is true Why???? SINGLE SAMPLE TEST, ONE-TAILED, 6 - Step Procedure for t Distributions, "one-tailed test" Step 1: Determine the hypothesis to be tested. Lower-Tail H0: μ ≥ μ0 H1: μ < μ0 or Upper-Tail H0: μ ≤ μ0 H1: μ > μ0 hypothesis test (lower or upper) = upper Step 2: Determine a planning value for α [level of significance] = 0.1 Step 3: From the sample data determine x-bar, s and n; then compute Standardized Test Statistic: t = (x-bar - μ0)/(s/SQRT(n)) x-bar: Estimate of the Population Mean (statistical mean of the sample) = 3.62 n: number of individuals in the sample = 40 s: sample standard deviation = 0.17 μ0: Population Mean = 3.54 significant digits = 3 Standardized Test Statistic t = ( 3.62 - 3.54 )/( 0.17 / SQRT( 40 )) = 2.976 Step 4: Use Students t distribution, 'lookup' the area to the left of t (if lower-tail test) or to the right of t (if upper-tail test) using Students t distribution Table or Excel TDIST(x, n-1 degrees_freedom, 1 tail) =TDIST( 2.976 , 39 , 1 ) Step 5: Area in Step 4 is equal to P value [based on n -1 = 39 df (degrees of freedom)] = 0.002 Table look-up value shows area under the 39 df curve to the right of t = 2.976 is (approx) probability = 0.002 Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 90% confidence. Conclusion: H1 is true Note: level of significance [α] is the maximum level of risk an experimenter is willing to take in making a "reject H0" or "conclude H1" conclusion (i.e. it is the maximum risk in making a Type I error).
Statistics testing the null hypothesis?
A sample study was made of the number of business lunches that executives claim as deductible expenses per month. If 40 executives in the insurance industry averaged 9.1 such deductions with a standard deviation of 1.9 in a given month, while 50 bank executives averaged 8.0 with a standard deviation of 2.1, test the null hypothesis µ1 - µ2 = 0 against the alternative µ1 - µ2 does not = 0 at a 5% level of significance. i know that n1 = 40, xbar1 = 9.1 and s1 = 1.9; and n2 = 50, xbar2 = 8.0 and s2 = 2.1 but what do I do next?? PLEASE HELP!
Statistics Hypothesis tail test question?
I know how to do the tail test but i'm not sure if its two tail test, left test or right test?? thank You An Auditor far a goverment agency is assigned the task of evaluating reimbursment for office visits to physicians paid by medicare. the audit was conducted on a sample of reimbursements. the same data are: Reimbursement [dollars] 31.1 81.84 111.3 110.65 121.69 76.59 52.5 73.24 27.59 105.09 69.84 57.02 108.04 136.99 143.71 95.16 138.05 129.82 124.3 72.61 Is there any evidence that the average reimbursment in the population is exactly $120? Test this hypothesis at the 1% 5% 10% level of significance, respectively.
Hypothesis Testing Part 1/2 Please Help me?
2. ANSWER: At a level of significance = 0.05 Conclusion: H1 is true Why???? SINGLE SAMPLE TEST, ONE-TAILED, 6 - Step Procedure for t Distributions, "one-tailed test" Step 1: Determine the hypothesis to be tested. Lower-Tail H0: μ ≥ μ0 H1: μ < μ0 or Upper-Tail H0: μ ≤ μ0 H1: μ > μ0 hypothesis test (lower or upper) = upper Step 2: Determine a planning value for α [level of significance] = 0.05 Step 3: From the sample data determine x-bar, s and n; then compute Standardized Test Statistic: t = (x-bar - μ0)/(s/SQRT(n)) x-bar: Estimate of the Population Mean (mean of the sample) = 2.4 n: number of individuals in the sample = 35 s: sample standard deviation = 0.29 μ0: Population Mean = 2.3 significant digits = 3 Standardized Test Statistic t = ( 2.4 - 2.3 )/( 0.29 / SQRT( 35 )) = 2.04 Step 4: Use Students t distribution, 'lookup' the area to the left of t (if lower-tail test) or to the right of t (if upper-tail test) using Students t distribution Table or Excel TDIST(x, n-1 degrees_freedom, 1 tail) =TDIST( 2.04 , 34 , 1 ) Step 5: Area in Step 4 is equal to P value [based on n -1 = 34 df (degrees of freedom)] = 0.025 Table look-up value shows area under the 34 df curve to the right of t = 2.04 is (approx) probability = 0.025 Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 95% confidence. Conclusion: H1 is true Note: level of significance [α] is the maximum level of risk an experimenter is willing to take in making a "reject H0" or "conclude H1" conclusion (i.e. it is the maximum risk in making a Type I error).
A sample of n = 25 scores produces a t statistic of t = 2.05. If the researcher is using a two-tailed test..?
Using this table. http://www.sjsu.edu/faculty/gerstman/Sta... Read row DF = 24 entries for two-tailed test. For DF = n - 1 = 25 -1 = 24 α = 0.1 , t = 1.711 (so it is better than this, so you can reject the NULL HYpothesis for α = 0.1 2.05 > 1.711 α = 0.05 t = 2.064 Since 2.064> 2.05 (the test statistic , you cannot reject the NULL hypothesis for α = 0.05 and of course , if you cannot reject for α = 0.05 , you cannot reject for any alpha smaller than 0.05 such as α =0.01 so based on this C. The researcher must fail to reject the null hypothesis with either α = .05 or α = .01.
Is it harder to reject the null hypothesis when conducting a two-tailed test rather than a one-tailed test? Wh
Without the aid of pictures, this will be long-winded... First off, keep in mind that hypothesis testing is about assuming something is true (the null hypothesis) until there is sufficient evidence (a surprisingly extreme test statistic) that we can reject the null hypothesis. What is 'sufficient evidence' is determined by 1. The level of the test, usually labeled with the Greek letter 'alpha' and often set at 5%. 2. If the test is one-sided or two-sided. I'll assume this is a Z or T test since this is probably the context in which this question arose. Right now only be concerned with the second aspect above regarding whether the test is one-sided or two-sided. Additionally, I'll refer to the following hypothesis tests as an example: One-sided: H_0: mu = 3 H_A: mu > 3 Two-sided: H_0: mu = 3 H_A: mu not= 3 With a one-sided test, we expect the test statistic to show up only on one side of the value in the null hypothesis. In the example above, we would expect our test statistic (Z or T) to show up above 0*, in which case we put our 'rejection region' only on this one side of 0. (Our 'rejection region' is the range of values our test statistic could take if we were to reject the null hypothesis.) * the mean is expected to be above 3, and Z or T will then be above 0. If the test is two sided, then we must split our rejection region between both sides (in the example, half goes above and half goes below). In such a case, the test statistic must be even more extreme since there is less of a chance if we were just looking at one side of 0. That is, Z or T must be even further from 0, but now it can be above or below 0, just further away. That is, our test statistic must be even further from 0. Since our test statistic must be more extreme in a two-sided test to reject the null hypothesis, it is harder to reject when the test is two-sided.
When do I use greater or less than signs in hypothesis testing?
Edit: I didn’t look very hard at the details to your question. I’m almost certain that you should use a one tailed test. So don’t find 2P, you want a single tailed test. It is up to you to decide which tail should lead to rejection. A discussion of this is in my original answer below.Original answer:The question is not completely specified. That’s not your fault, that’s the way the world is. You have to use your common sense to fill in the details. Do you think that female students think that the average is not $95, but could be more or less? I doubt it. This would be the two sided hypothesis (which is sometimes useful) [math]H_0: mu = 95[/math] against [math]H_1: mu \ne 95[/math]. That would be ridiculous in this case.So either they think it is less than $95 on average, or they think it’s more than $95 on average. Which do you think they are more likely to believe? That becomes the alternative hypothesis. The null hypothesis is the opposite. They expect the data to reject the null hypothesis so they will have proved their point. But until they do the test they don’t know for sure. (They don’t after the test either, but they can quantify the amount of evidence.)So (you decide), they either test [math]H_0: mu \le 95[/math] against [math]H_1: mu > 95[/math] or they test [math]H_0: mu \ge 95[/math] against [math]H_1: mu < 95[/math].Actually, there is a clue in the given data: the sample mean is actually $110, but it is cheating to use that; you should frame your hypotheses before collecting the data.The actual test is to see if 110 is not just greater than 95, but so much greater that chance alone is not likely to account for difference. (There, I’ve told you which hypothesis is which—if the hypotheses were the other way round, the data would support the null hypothesis without you needing to do any calculations.)In practice, we usually write the null hypothesis as [math]H_0: mu = 95[/math] because if you can reject this at the 5% level (or 1%, or whatever) then you would automatically reject any smaller value at an even smaller level.
Statistics Question Part 2?
A hypothesis test with a sample of n = 25 participants produces a t statistic of t = 2.53. Assuming a two-tailed test, what would be the correct statistical decision? 1. The researcher must fail to reject the null hypothesis with either α = .05 or α = .01. 2. It is impossible to make a decision about H0 without more information. 3. The researcher can reject the null hypothesis with α = .05 but not with α = .01.