Because 1818 does not fall in the rejection region H is NOT REJECTED at the 01

Because 1818 does not fall in the rejection region h

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Because -1.818 does not fall in the rejection region, H 0 is NOT REJECTED at the .01 significance level. We have NOT demonstrated that the cost-cutting measures reduced the mean cost per claim to less than $60. The difference of $3.58 ($56.42 - $60) between the sample mean and the population mean could be due to sampling error. Step 4: Formulate the decision rule. Reject H 0 if t < -t D ,n-1 343 NOTE: Calculation discrepancy in z-value
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The current rate for producing 5 amp fuses at Neary Electric Co. is 250 per hour. A new machine has been purchased and installed that, according to the supplier, will increase the production rate. A sample of 10 randomly selected hours from last month revealed the mean hourly production on the new machine was 256 units, with a sample standard deviation of 6 per hour. At the .05 significance level can Neary conclude that the new machine is faster? Testing for a Population Mean with an Unknown Population Standard Deviation- Example Step 1: State the null and the alternate hypothesis. H 0 : μ £ 250 H 1 : μ > 250 Step 2: Select the level of significance. It is .05. Step 3: Find a test statistic. Use the t distribution because the population standard deviation is not known and the sample size is less than 30. Testing for a Population Mean with an Unknown Population Standard Deviation- Example
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Step 4: State the decision rule. There are 10 – 1 = 9 degrees of freedom. The null hypothesis is rejected if t > 1.833. Step 5: Make a decision and interpret the results. The null hypothesis is rejected. The mean number produced is more than 250 per hour. 162 . 3 10 6 250 256 ! 0 ! 0 ! n s X t P Testing for a Population Mean with an Unknown Population Standard Deviation- Example Tests Concerning Proportion z A Proportion is the fraction or percentage that indicates the part of the population or sample having a particular trait of interest. z The sample proportion is denoted by p and is found by x/n z The test statistic is computed as follows: 350
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Assumptions in Testing a Population Proportion using the z-Distribution z A random sample is chosen from the population. z It is assumed that the binomial assumptions discussed in Chapter 6 are met: (1) the sample data collected are the result of counts; (2) the outcome of an experiment is classified into one of two mutually exclusive categories—a “success” or a “failure”; (3) the probability of a success is the same for each trial; (4) the trials are independent z The test we will conduct shortly is appropriate when both n S and n (1- S ) are at least 5. z When the above conditions are met, the normal distribution can be used as an approximation to the binomial distribution 349 Assumptions when using a Normal approximation for hypothesis testing with a proportion ( short version ) 1. It meets the conditions of a BINOMIAL distribution 2.
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