Hypothesis testing steps?
One way is to observe on a number of rude cases, and calculate the proportions who are rude to to fellow drivers who drive low status cars (p1) and to those who drive high status cars (p2). Test of difference in proportions You are testing about the proportions between the two groups. H0: θ1= θ2 HA: θ1> θ2 (Higher proportion towards low status) The test statistic has a value z = (p1 - p2) / (estimate of standard error of difference), with p1 = n1/(n1+n2), p2 = n2/(n1+n2); n1 = no. of rudes for low, n2 = no. of rudes for high. Total no. of "rudes" you have observed is n1+n2. Std error = √ {θ(1-θ)(1/n1 + 1/n2)}. Under H0, θ is estimated by {(n1)(p1) + (n2)(p2)} / (n1+n2). One-tailed test, right, critical region on right tail. Level of Significance: say 0.05 Decision Rule: Reject if p-value < 0.05 Decision: Reject Ho at 0.05. The claim is supported.
Testing claims about proportions - statistics help?
For this problem I have to identify the null hypothesis, alternative hypothesis, test statistic, p-value or critical value, conclusion about null hypothesis and final conclusion that adresses the original claim. In a Pew Research Center of 745 randomly selected adults, 589 said that is morally wrong not to report all income or tax returns. use a 0.01 significance level to test the claim that 75 % of adults say that it is morally wrong to not report all income on tax returns. So far I got the hypothesises. H0= μ= .75 h1= μ doesnt = .75 Not sure where to go from here... help please?
How do I make the null hypothesis in a hypothesis test?
The null and alternative hypothesis are mutually exclusive and exhaustive so that one is true to the exclusion of the other. Also, null hypothesis always contains the ‘=’ sign (may be of the form '=' or '<=' or '>=') while alternative hypothesis never contains the '=' sign (may be of the form '!=' or '>' or '<' respectively) .And the steps for writing the hypotheses areas follows:Look for the claim in the research question.The claim can be written as null or alternate hypothesis on the basis of presence of equal sign in it.State the other hypothesis.Few more examples are: A consumer analyst reports that the mean life of a certain type of automobile battery is not (claim) 74 months.H0 : mean = 74H1: mean ≠ 74 A radio station publicizes that its proportion of the local listing audience is greater than or equal to (claim) 39 %H0 : proportion ≥ 39H1: proportion < 39 Suppose we wanted to test the hypothesis that the mean familiarity rating exceeds 4.0, the neutral value on a 7 point scale.H0 : mean ≤ 4.0H1: mean > 4.0
Find the P-value for the indicated hypothesis test.?
Again, I am so sorry that I have made mistakes in my earlier answer. The following hypothesis is a one-sided- NOT two sided as I perviously stated. n = sample size = 130, p = p0= the true proportion of students go into general practice = 28% = 0.28 Since np = 130 * 0.28 and n(1-p) = 130 * (1-0.28) np = 36.4 = 93.6 are both greater than 5, so the sample size is large and we can use the t distribution to test the hypothesis. We then set up the following set of one sided hypothesis to be tested: Ho: p < = p0 versus Ha: p > p0 H0: p < = 28% versus Ha: p> 28 % H0: p < = 0.28 p > 0.28 since the medical school hopes its claim is true- that is more than(>) 28% go to general practice. z = test static = (sample proportion - p0)/ [√p0(1- p0)/n] z = (0.32 - 0.28)/([√0.28 (1- 0.28)/ 130] z = 1.015749 The p value is equal to the right end area of z = 1.015749 in the normal distibuted curve respectively = (0.5- 0.3461*) = 0.5139 * you can check the right end area of z= 1.015749 from a Table of areas in the standard normal curve.
What are some examples I could use for a hypothesis test project for statistics?
I'm not exactly sure what your project is asking, but for my Statistic's project I used a Chi-Square test for independence. The test allows you to see if there is an association between the two variables you look at. My study looked at whether or not there is an association between the number of siblings a student has and that student’s performance in school, relative to their approximate grade point average. What we did was polled the students at my school (picked by assigning all students a number and then picking them with a random number generator). To do a Chi-Squared test, you must make sure that the necessary conditions are met: 1) That the sample was taken from an SRS 2) That there are not expected counts less than 1 3) That no more than 20% of the expected counts are less than 5 You then need to come up with a null and alternative hypothesis, and then pick an alpha level. You then can proceed with your Chi-square test. I hope this helped. Good luck!
One sided confidence interval for a test of proportion.?
A random sample of size 500 connecting rod pins contains 65 nonconforming units. Determine if it's possible that the average defective is more than 0.13 with 95% confidence. I understand that I could solve this using a hypothesis test or confidence intervals. I'd like to solve this using confidence intervals. I'm struggling with the actual question that my professor came up with. By using the formula p < or = p-hat + z*sqrt((p-hat(1-p-hat))/n) I determined that p < = 0.1551. I thought that this meant that this was the upper limit of the confidence interval. So to me, that means that my upper limit of the 95% confidence interval is 0.1551 and that it's possible for the average fraction defective to be more than 0.13 at a confidence interval of 95%. What I don't get is that 0.13 is p-hat, so.... shouldn't it always be possible for the average fraction defective to be more than 0.13? Now if he had asked "determine if it is possible that the average fraction defective is more than 0.19..." then this question would make more sense to me because I would say, "no, not at the 95% confidence level because the limit is 0.1551". My professor corrected my answer and wrote that I have to put p > = to determine if it's greater than 0.13, but that doesn't make any sense to me. It does make sense to me if I was running a hypothesis test and trying to determine the probability of the fraction defective to be greater than 0.13. Then I was just be trying to find 1-P(p-hat < = 0.13). Thanks in advance.
What statistical test should be used for an A/B test on conversion rates?
Yes, we use Z-test (although for Chi-square test for two binomial samples) would be equivalent. Binomial can be approximated as a normal distribution, and after that it is a simple matter of applying a two-tailed hypothesis test. As I said, you can also apply a Chi-square test, it will give the exact same p-value (although chi-square test does not give confidence intervals).Here are some math details: http://visualwebsiteoptimizer.co...
Statistics: Calculations for testing claims Need Help Please?
Large Sample Hypothesis Test for the Difference in Proportions Let X be the number of success in nx independent and identically distributed Bernoulli trials, i.e., X ~ Binomial(nx, px) Let Y be the number of success in ny independent and identically distributed Bernoulli trials, i.e., Y ~ Binomial(ny, py) Let pxHat = X / nx pyHat = Y / ny pHat = (X + Y) / (ny + ny) Assuming that (nx + ny)*pHat > 10 and (nx + ny)*(1-pHat) > 10 (some will say the necessary condition here is > 5, I prefer this more conservative assumption so that the approximations in the tail of the distribution are more accurate), then to test the null hypothesis H0: px - py = Δ or H0: px - py = Δ or H0: px - py = Δ Find the test statistic z = ((pxHat - pyHat) - Δ ) / (sqrt(pHat * (1 - pHat) * (1/nx + 1/ny)) The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis. H1: px - py > Δ; p-value is the area to the right of z H1: px - py < Δ; p-value is the area to the left of z H1: px - py ≠ Δ; p-value is the area in the tails greater than |z| If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true. If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible. The hypothesis test in this question is: H0: px - py = 0 vs. H1: px - py ≠ 0 pxHat = 101 / 10239 = 0.009864245 pyHat = 56 / 9877 = 0.005669738 pHat = (101 + 56) / (10239 + 9877) = 0.007804733 The test statistic is: z = (( 0.009864245 - 0.005669738 ) - 0 ) / ( √ ( 0.007804733 * (1 - 0.007804733 ) * ( 1 / 10239 + 1 / 9877 ) z = 3.379666 The p-value = P( Z > |z| ) = P( Z < -3.379666 ) + P( Z > 3.379666 ) = 2 * P( Z < -3.379666 ) = 0.0007257394 Since the p-value is less than the significance level of 0.05 we reject the null hypothesis and conclude the alternate hypothesis px - py ≠ 0 is true.