Question 983247
First state the null hypothesis(Ho) and the alternative hypothesis(Ha),
Ho is average career in professional football is longer than 4.7 years
Ha is average career in professional football is less than or = 4.7 years 
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significance level is 0.05
the test method will be a 1-tailed test
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using sample data, we calculate the standard error(SE) of the mean, degrees of freedom(DF), t-score test statistic,
SE = s / square root(n), where s is the standard deviation of the population and n is the sample size
SE = 2.1 / square root(50) = 0.296984848
DF = n - 1 = 50 - 1 = 49
t = (sample mean - pop mean) / SE = (5.5 - 4.7) / 0.296984848 = 2.69374012
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we use t-distribution to determine probability(P) of t < 2.69374012, we find
P( t < 2.69374012) = 0.9952
This means we would expect to find a sample mean of 5.5 or smaller in 99.5 percent of our samples, if the true population mean were 4.7. Thus the P-value in this analysis is 0.9952
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Since the P-value (0.9952) is greater than the significance level (0.05), we cannot reject the null hypothesis.