The reason for this is because in tests like these observed differences are usually due to chance differences in sampling. Meaning that, let’s say the p value is more than the significant level of what is being tested or if it’s less than what is being tested. If it is more the null hypothesis is rejected, and if it is less the null hypothesis is rejected in favor of an additional hypothesis. Which is more in the line of saying that it is accepted. The null hypothesis for this experiment was that there would be no changes in allele or genotype frequencies.
False The F test assumes that the sampled populations are not normal. A. True B. False The F test for equality of two variances tests whether the ratio of the sample variances differ significantly from 1. A.
Do these scores strengthen or weaken the validity of the research results? Provide a rationale for your answer. The mean baseline and posttest depression scores of the control group are exactly the same at 10.40. This adds strength and validity to the research as it show that there is no difference in their depression as this group did not attend the empowerment program. 5.
Although the data are slightly right skewed, because of the large sample size(n=152) our inference on the mean is valid, by the Central Limit Therom. The Box Cox plot suggests a power transformation using a power, p= -0.5 would make the data more shaped like a normal distribution. The Normal Q-Q plot shows the transformed data points lying close to the straight line and the box plot of the transformed data looks symmetric with two outliers. The numerical summary of the transformed data confirms that the transformed data are symmetric (mean=1 and median=1). A Shapiro-Wilk test on the transformed data provides no evidence against the transformed data having come fro a normal distribution (P-value=0.4286).
1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
Secondary outcomes were disability and quality of life measures. 1. The major strengths were that the study eliminated the potential for a differential response bias by using hospital records and not patient self-reporting. It was a successful randomization
Chi Square Analysis Questions Based on our group’s sample, we should reject the null hypothesis. The null hypothesis states that the frequencies of each color M&Ms in our sample size of 5 bags would match exactly the frequencies given to us by the Mars Company (maker of M&M candies). Our data, however, showed that the frequencies of the M&Ms were drastically different (p-value less than 1%) Since we rejected the null hypothesis, some explanations that could have caused this is human error from the factory workers, the percentages may have changed since they were last checked, shortage of a certain dye, or a deviant employee stealing M&Ms of a certain color. Based on our class’s sample, we should reject the null hypothesis. The null hypothesis states that the frequencies of each color M&Ms in our sample size of 30 bags would match exactly the frequencies given to us by the Mars Company (maker of M&M candies).
Task 1: Solve the following problems: * A student of the author surveyed her friends and found that among 20 males, 4 smoke and among 30 female friends, 6 smoke. Give two reasons why these results should not be used for a hypothesis test of the claim that the proportions of male smokers and female smokers are equal. * Understand that this is not considered a random sample, but, a convenience sample. So you could not get the results that you would get from a random sample. * When setting up the formula you get np = x = 4 < 5, so the normal distribution cannot be used to get the approximate of binomial distribution * Given a simple random sample of men and a simple random sample of women, we want to use a 0.05 significance level to test the claim that the percentage of men who smoke is equal to the percentage of women who smoke.