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Topic: **Hypothesis on****Question:**
What Makes a Good Hypothesis? Yes/no
1) The hypothesis is based on information contained in the review of literature?
2) The hypothesis includes the independent and dependent variables?
3) You have worded the hypothesis so that it can be tested in the experiment?
4) If you are doing an engineering or programming project, have you established your design criteria?

June 16, 2019 / By Letitia

You are trying to establish criteria for good hypothesis based on these scenerios. I think there can not be firm yes/no answer for any of them. 1. The information contained in review of literature. There can not be a way to judge ALL reviews to be 100% accurate and unbiased or not. 2. Question of variables is again context dependent. We cant know the critical role of each variable in a general approach 3. This is certainly an ESSENTIAL requirement of any hypothesis. Otherwise the theory to be proved can be anything! 4. Design criteria are again essential for any engg project. This is what came to my mind when i looked at your list. If you have any idea about the nature of the problem you can discuss.

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ANSWER: Conclusion: H1 is true. The average number of days is less than 20. 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) = lower 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 (statistical mean of the sample) = 43.04 n: number of individuals in the sample = 50 s: sample standard deviation = 41.9 μ0: Population Mean = 20 significant digits = 2 Standardized Test Statistic t = ( 43.04 - 20 )/( 41.9 / SQRT( 50 )) = 3.89 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( 3.89 , 49 , 1 ) Step 5: Area in Step 4 is equal to P value [based on n -1 = 49 df (degrees of freedom)] = 0 Table look-up value shows area under the 49 df curve to the left of t = 3.89 is (approx) probability = 0 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).

well you simply just leave the question blank and then drop out of university, then marry earl and live on the roof of montannas. JWANNNNNNNNNA.

You have 15 pairs of numbers: (X_1, Y_1), (X_2, Y_2), ..., (X_15, Y_15) You calculate the average of (Y_k-X_k). That is (1/15)*Sum(Y_k-X_k). Presumably this average is greater than 0. You want to know if it is likely that, for the ENTIRE population, from which your sample is drawn, the average difference is greater than 0. (Are the Y's generally bigger than the X's?) Your "null hypothesis" is that this average difference is zero. You know that for randomly choosen samples, the distribution of the averages is Normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of 15. (Actually this is for an infinite population. For a finite population the distribution is binomial, which is approximately normal for large populations.) So, assuming the "null hypothesis", the mean of YOUR sample has a 99% chance of being smaller than 2.33 times the standard deviation of the population divided by the square root of 15. Unfortunately, you do not know the standard deviation of the population. Worse, on average, the standard deviation of your sample does not even match that of the population. Calculate a modified standard deviation for your sample (called the "sample standard deviation" in most references: SqareRoot(Sum(((Y_k-X_k)-Mean)^2/14)) Note the "14" instead of "15". This, on average, matches the population standard deviation. [The proof is too long to insert here, but can be done by a good high school student. I have had my students do it in the past.] Unfortunately, that modified standard deviation only matches "on average". So, the distribution of your sample mean divided by this modified standard deviation is still not normal. That distribution does matches the "Student's T" distribution, with 14 degrees of freedom. (Not 15 degrees of freedom!) Now you can use that distribution to calculate the probability that this "T" statistic (the sample mean minus the population mean divided by the modified standard deviation over squareroot(15)) is greater than some given quantity. Use the Student's T tables from your text, or the TINV function in your favorite spreadsheet program. Note: The TINV(probability, Degrees of freedom) function takes a probability and gives a number for which the probability is the probability that the absolute value of the T statistic is greater than that number. It appears you are interested in the signed value, not the absolute value, so double the probability: =TINV(2*0.01, 14) = 2.624 (1% chance of greater than 2.624 and 1% chance of less than -2.624) Assuming the null hypothesis, there is a 99% chance that the T statistic will be less than 2.624. If it is greater than this amount you can reject the null hypothesis. (There is also a 99% chance that the T statistic will have an absolute value less than 2.977=TINV(0.01,14), but I think you only care about positive T statistics. - I have no idea where 2.602 might come from...) Does this help?

For 5% significance: >> I think that you reject the null hypothesis when the generated p value is less than 0.05, and you accept the null hypothesis when the generated p value is greater than 0.05? Correct. For a one-tail chi-square test: If the calculated chi-square value is greater than the critical value, reject the null hypothesis. The two tests give the same result.

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