Question: Find the test statistic. 2.5/ 100 = 0.25 Z = 2.99 - 2.983 = 0.007 / 0.25 = 0.028 The test statistic (z) is approximately

Find the test statistic. 2.5/ 100 = 0.25 Z = 2.99 - 2.983 = 0.007 / 0.25 = 0.028 The test statistic (z) is approximately 0.028. c. Find the pvalue. .0175 d. Describe what the pvalue tells you about this situation. Use a sketch in your description. A p value of 0.0175 indicates that there is a 1.75% chance of observing an average TV watching time as extreme as 2.983 hours (or more extreme) if the true average TV watching time for BC students is indeed 2.983 hours. e. Does this provide evidence that BC students really have a higher average than the nation

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