#### Answer

There is sufficient evidence to support that the common beleif is wrong.

#### Work Step by Step

$H_{0}:\mu=98.6.$ $H_{a}:\mu \ne98.6.$ Hence the value of the test statistic: $\frac{\overline{x}-\mu}{s/\sqrt n}=\frac{98.2-98.6}{0.62/\sqrt{106}}=-6.64.$ The P-value is the interval of probabilities between which the value of the test-statistic lies in the table with degree of freedom=sample size-1=106-1=105, hence P is less than 0.01. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P is less than $\alpha=0.05$, because it is less than 0.01, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that the common beleif is wrong.