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Null and Alternative Hypotheses for a Mean

Here is a numerical example: You need both tails probabilities for test of hypothesis and construction of confidence interval for the ratio of two independent populations' variances.

The test statistic for examining hypotheses about one population mean:

A study is designed to test whether there is a difference in mean daily calcium intake in adults with normal bone density, adults with osteopenia (a low bone density which may lead to osteoporosis) and adults with osteoporosis. Adults 60 years of age with normal bone density, osteopenia and osteoporosis are selected at random from hospital records and invited to participate in the study. Each participant's daily calcium intake is measured based on reported food intake and supplements. The data are shown below.

Support or Reject Null Hypothesis

Click the link the skip to the situation you need to support or reject null hypothesis for:

In other words, the simplest correction is to move the cut-off point for the continuous distribution from the observed value of the discrete distribution to midway between that and the next value in the direction of the null hypothesis expectation.

Following the above process for this test, the K-S statistic is 0.421 with the p-value of 0.0009, indicating a strong evidence against the null hypothesis.

How to Set Up a Hypothesis Test: Null versus Alternative

We are testing the hypothesis that the population means are equalfor the two samples.

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

Broken down into English, that’s H0 (The null hypothesis): μ (the average) = (is equal to) 8.2
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Null hypothesis: μ = 72 Alternative hypothesis: μ ≠72

The following chart summarizes application of statistical tables with respect to test of hypotheses and construction of confidence intervals for mean and variance in one population or the comparison of two or more populations.

Null hypothesis: μ = 72 Alternative hypothesis: μ ≠72

Note also that, in hypothesis testing concerning the parameter of binomial and Poisson distributions for large sample sizes, the standard deviation is known under the null hypotheses.

We reject the null hypothesis because -6.15

As you know by now, in test of hypotheses concerning , and construction of confidence interval for it, we start with known, since the critical value (and the p-value) of the Z-Table distribution can be used.

Then, we hypothesize that the 's are equal wich is null hypothesis.

Given xbar = 492 construct a 95% confidence interval for given = 16 Plugging in the numerical values, one gets:
P[476.3 507.7] 0.95 Notice the Duality between the test of hypothesis and confidence interval.

If the null hypothesis is true, then the mean is 0.

Are there any connections between test of hypotheses and confidence interval under different scenario, for example testing with respect to one, two more than two populations?

Know that, for any test of hypothesis, there is only one p-value.

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

For example, using Z-test of hypothesis in the following Figure.

Central Limit Theorem (CLT) If E(X) = , Var(X) = , then for large n, say, n 30As a strong result, the CLT implies that if the sample size is large enough, then one may relax the normality condition whenever dealing with the question of testing or constructing confidence interval for population’s mean ().

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