quiz 5 stat - 1of10ID:MST.HT.TM.03.0010 . .Afootnoteatthe

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 1 of 10   ID: MST.HT.TM.03.0010 Geoff is the proud owner of a restaurant. Recently he read an article in  Food Weekly  which stated that the  population mean preparation time for food at popular restaurants is equal to 28 minutes. A footnote at the  bottom of the article said that the population standard deviation of preparation times was equal to 6 minutes. Geoff is interested in determining whether his restaurant has the same average preparation time as the  popular restaurants described in the magazine article. Geoff timed how long it took to prepare 39 randomly  selected meals over a week. The mean preparation time for the sample was calculated as 27minutes. Geoff  would like to use his sample to construct a hypothesis test with H 0 μ  = 28 and H a μ    28. a) Calculate the  P -value that corresponds to the sample and hypotheses. You may find this  standard normal   table  useful. Give your answer as a decimal to 4 decimal places. P -value =  b) Using the  P -value for Geoff's hypothesis test and a level   = 0.01, Geoff should α   the null hypothesis. [1 out of 2] -    Feedback a) This is not correct. P -value =  0.2980 b) You are correct. Calculation a) The  P -value measures the probability of obtaining a sample mean equal to (or more extreme) than the  calculated sample mean. This probability can be obtained by calculating the test statistic that corresponds to  the null hypothesis for the mean and then using the  standard normal table  to obtain the probability of  obtaining such a test statistic (or one more extreme). The test statistic can be calculated using the following formula:  show variables z = x - μ 0 σ n = 27 - 2 8 6 39
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= -1.040833... = - 1.04 Rounded as last step The  P -value can now be calculated as the probability that |z| > -1.04. That is, the  P -value for a two-tailed  hypothesis test is the probability that the calculated sample mean is extremely large (z > 1.04) plus the  probability that the calculated sample mean is extremely small (z < -1.04). Since the standardized normal  distribution is symmetric, these two probabilities are equal. Therefore, the  P -value can simply be calculated  as2  ×  Probability(z > 1.04). According to the  standard normal table , this probability is equal to 2  ×  0.1490 =  0.2980 . b) The null hypothesis is rejected if the  P -value is less than the level of significance of the test. This is  because the level of significance defines the standard by which statisticians decide that there is enough  evidence to suggest that the null hypothesis is false. The level of significance of this test is   = 0.01. α According to the  P
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