Significance Level for Cereal Consumption

Related BrainMass Solutions

• Analyzing the given demand function

  lnQ=9.01-0.68lnP+0.75 lnA-1.3 lnM, Where is the quantity of cereal ,P is the price of Cereal , A is the level of advertising and M is the income .

• Writing the Null and Alternative Hypothesis

  Level of Significance: α = for the given level of significance Decision rule: If the P-value of the test statistic is less than 0.05 we reject the null hypothesis. Otherwise accept the null hypothesis.

• Hypothesis

  Perform a hypothesis test at the 5% significance level.

• Hypothesis Testing for Filling Variance of Cereal Boxes

  Conclusion: Since the calculated test statistic value is less than the critical value we may accept the null hypothesis at 0.01 level of significance.

• Hypothesis Testing of Varaiance: Cereal box filling weight

  = 51.2 Rejection Criteria: Reject the null hypothesis, if the calculated value of chi-square statistic is greater than the critical value of chi-square at the 0.05 significance level.

• Statistics - Hypothesis Test

  A sample of 100 boxes of cereal is taken from the packaging line. The mean of samples is 11.85 oz., with a standard deviation of 0.5 oz. The level of significance is 0.05. 1) First, we define our null hypothesis.

• Hypothesis Testing, Significant Levels and p-Values

  At the .05 significance level, test the claim that the mean waiting time is less than 10 minutes. 7. A cereal company claims that the mean weight of the cereal in its packets is more than 14 oz.

• Test of Hypothesis Using T test

  Use a 0.05 level of significance.

• Hypothesis Testing Problems: Value and Traditional Approaches

  Find x-bar and s for this data set. b. At the 5% significance level, test the claim that for the modified components, the mean time between failures is greater than 520 hours?