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Solution 3 of 4 B8 Tables A publisher reports that 76 % of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 72 % of the readers owned a particular make of car. Is there sufficient evidence at the 0.01 level to support the executive's claim? Now that we have the test statistic, we can calculate the P-value. The P-value is the probability of observing a value of the test statistic as extreme or more extreme than the one observed in the data, assuming the null hypothesis is true. In this situtation, that means that the value we want to find is P(|z] > 1.32). Using either the standard normal tables or technology, we can find this value as follows. 0.0934 093417 -1.32 0 1.32 P(]z] = 1.32) = 0.093417 + 0.093417 ~ 0.1868 Thus the P-value of the test statistic is 0.1868. Continued on the next page ...