Answer:
The claim is valid for 98% confidence interval. A further explanation is described below.
Step-by-step explanation:
The given values are:
Sample size,
n = 38
Sample mean,
[tex]\bar{x}=32700[/tex]
Population standard deviation,
[tex]\sigma=8740[/tex]
For 98% confidence interval,
Mean = [tex](\bar{x} \pm z^*\times \frac{\sigma}{\sqrt{n} } )[/tex]
On substituting the given values, we get
= [tex](32700 \pm 2.326\times \frac{8740}{\sqrt{38} } )[/tex]
= [tex](32700 \pm 3297.84)[/tex]
= [tex](29402.16,35997.84)[/tex]
Thus the above is the appropriate solution.
