Answer
There is sufficient evidence to support that the diet is effective.
Work Step by Step
$H_{0}:\mu=0.$ $H_{a}:\mu>0.$ Hence the value of the test statistic: $\frac{\overline{x}-\mu}{s/\sqrt n}=\frac{3-0}{4.9/\sqrt{40}}=3.872.$ 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=40-1=39, 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 diet is effective.
