Answer:
yes, he was effectively a .400 hitter
Step-by-step explanation:
The mean rate of success ( ÿ ) is 0.394.
The total number of hits was 419.
Under table t the value at 5% confidence interval is 1.96.
Here se the asymptotic standard deviation is given as follows
se = [tex]\sqrt{\frac{y(1 - y)}{n} }[/tex]
= [tex]\sqrt{\frac{0.394(1 - 0.394)}{419} }[/tex]
= 0.02387
The confidence interval is calculated as follows:
y ± 1.96 * se
0.394 ± 1.96 * 0.02387
the answer is -0.347 or 0.4407)
meaning that the mean lies between -0.347 to 0.4407.
There is strong evidence that 0.4 lies between the above confidence interval.
This means that Gwynn was effectively a 0.4 hitter
