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H 0 : A 1 = A 2 = . . . = A = 0  Field 1: Tukey, HSD, or Tukey's  Field 2: means or mean  to be normally distributed.  to have a zero mean and constant variance. used to control for potential confounding influences. that the factor effects do not differ significantly from zero.  repeated treatment observations can be impossible to collect.  the observations are expensive. y ij = μ + A + B j +ε ij  Field 1: block  Field 2: treatments or treatment   Field 1: SSA  Field 2: SSB  Field 3: SSE  Field 1: block

 Field 2: treatments or treatment fixed-effects model  Field 1: high or large  Field 2: null A random sample of 10 coffee orders at a café shows an average preparation time of 3.5 minutes with a standard deviation of 0.8 minutes. You want to test the significance of the true mean preparation time using a t-test. What is the decision rule in this case? Is there anyone who can answer this question? You get one response credit for discussion; however, the response should be at least 15/20 words or longer. I look forward to reading your answers. Also, please remember to respond during the week to two of your team member's posts so that you can earn full participation points. To determine the decision rule for the t-test, you need to establish the significance level (alpha) and the degrees of freedom. Then, compare the calculated t-value with the critical t-value from the t-distribution table. If the calculated t-value is greater than the critical t-value, you reject the null hypothesis. If it's less, you fail to reject the null hypothesis. Summarizing the importance of the decision rule in testing the significance of the true mean coffee preparation time. Emphasizing the need for statistical analysis to make informed decisions based on data. Since you have a sample size of 10 coffee orders, the degrees of freedom for the T-test would be (n-1), which is 9 in this case. For a two-tailed T-test with a significance level (alpha) of 0.05, the critical values are determined

by looking up the T-distribution table or using a statistical software. For a sample size of 10 and a two-tailed test at a 0.05 significance level, the critical values would be approximately -2.262 and 2.262. So, if the calculated T-statistic falls outside of this range, we would reject the null hypothesis. Hope that helps! Let me know if you have any other questions.

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