Chapter 11: Multiple Comparison and Analysis of Variance or Anova

Chapter 11: Multiple Comparison and Analysis of Variance or Anova

Test the hypothesis that the mean number of hours studying varies by major

1. Set up the hypothesis

Since the problem is about the MEAN, the test is about a single population mean.

Therefore set the null and the alternative hypothesis:

In this case, there is an implied challenge or claim. Thus the study is to calculate the number of hours varied by the three majors to carry out the research. The main effect of this study is to set the hypothesis as a one-tailed test. The claim will be in the alternative theory as a one-tailed test.  Therefore the application will always be in the alternative hypothesis because the burden proof still lies with the alternative. Moreover, the status quo will be defeated with a high degree of confidence. In this case, 95% is the rate of confidence. The null and alternative hypotheses are thus

H₀: μ ≥ 16.45

H_a: μ < 16.45

1. Calculator: write down clearly how you input data in the calculator.
2. Go to 2: T-Test
3. Make sure the input is set to Stats, Not Data. This is because one is dealing with raw data. (In this case, the raw data us the calculated mean between the three majors.)
4. Enter the data for the test statistic to be able to compute.
   1. Is the hypothesized mean under the null.
   2. the observed average of the sample. ( the three majors subjects)
   3. is the SD+ of the example.
   4. The sample size.
5. Be specific on the alternative hypothesis. The average of the box or
6. Calculate the results of the test

1. Write down the p-value from the calculator.

To use the p-value system from the calculator to conclude, the calculation could be the statistic that is taken as an additional step to find the probability of being 2.08. The standard deviations from the mean on a t-distribution the value are .0187. Therefore, comparing this to the α-level of 0.05, the null cannot be accepted. Because p < α, therefore, the null hypothesis is rejected.

H0 : μ = 40

Ha : μ > 40

p = 0.0062

1. Decide to reject the null hypothesis if the significant level is 0.05

The null hypothesis cannot be accepted since the data supports the claim that the average conductivity is greater than one if the significant level is 0.05.

1. What does it mean it means to reject the null hypothesis.

As described in the task given here on testing the hypothesis p <0. 05, the rejected null hypothesis is defined according to the given context that an average of the dependent variable is not the same for all the majors offered for the students. With ANOVA, if the null hypothesis is rejected, then at least two groups are different, and in this case, the majors given here are not the same. The null hypothesis in ANOVA is always that there is no difference in means. The research hypothesis captures any difference in ways and includes, for example, (the situation where all four ways are unequal, where one is different from the other three, two are different, and so on.) The alternative hypothesis, as shown above, captures all possible situations other than equality of all means specified in the null hypothesis. Therefore after the hypothesis test has taken place, there are only two possible outcomes. When your p-value is less than or equal to your significance level, you reject the null hypothesis. The results are statistically significant. In this case, the p-value is less considerable than the significance. Therefore you must reject the null hypothesis