Answer:
The correct option is;
Yes because the computed t lies in the rejection region
Step-by-step explanation:
Here we have
The calculated mean of the values given as
[tex]\bar x[/tex] = Sample mean = 41.5
μ = Population mean = 43
σ = Standard deviation 1.784
The confidence interval is
[tex]CI=\bar{x}\pm t\frac{s}{\sqrt{n}}[/tex]
At 0.02 the critical t is [tex]\pm[/tex] 2.718 with a p value of 0.014 which is less than the confidence level at 0.02. We therefore reject the null hypothesis and we fail to reject the alternate hypothesis which states that there has been a change in the mean length
Therefore the answer is Yes because the computed t lies in the rejection region.