First question, How to set assumtion... Third , 5-1 = 4 degrees of freedom .. Fo

Question

First question, How to set assumtion... Third ,  5-1 = 4 degrees of freedom .. Fouth.. Decision part.. Could you explain step by step overall. I do want to learn sohelp me out^^@~ thank you in advance.. ---------------------------------------------------------------------------------------------------- The specifications for a certain kind of ribbon call for amean breaking strength of 180 pounds. if five spieces of the ribbon(randomly selected from different rolls) have a mean breakingstrength of 169.5 with a standard deviation of 5.7pounds, test the nullhypothesis =180 pounds against the alternative hypothesis < 180 pounds at the 0.01 level of significance.Assume that the population distribution is normal. ---------SOLUTION------------- 1. null hypothesis : = 180 pounds      alternative hypothesis : <180 pounds 2. Level of significance : = 0.01 3. Criterion : reject the null hypothesis if t < -3.747,where 3.747 is the value of 5. Decision: Since t = -4.12 is less than -3.747, the nullhypothesis must be rejected at level = 0.01    in other words, the breaking strength is belowspecifications. the exact tail probability, or P value, cannot bedeterminde from Table 4, but it is 0.0073. the evidence against themean breaking strength being 180 pounds is even stronger than 0.01.only about 7 in 1000times would we observe a value of t that is-4.12 or smaller , if the mean really were 180 pound..                   

Explanation / Answer

In a sample size of n = 5, the df is n-1 = 4. Look up the critical t corresponding to df = 4 and = 0.01. Remember that you have a 1-tailed test. You want the area to the left to be 0.01. The t-distribution looks like a normal distribution in that itis symmetrical about 0. So your critical t value is negative, because of the directionof the alternative hypothesis. If your altenative were:        H1:>180 then the critical t would be positive 3.747. In your case, the critical t is -3.747. The critical region is-3.747 or less. Since the test statistic that you calculate, -4.12, lies inthe critical region, you would reject the null. Hope this helps, Mike

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