Elementary Statistics (12th Edition)

(1)$\mu$ is between 0.6866 and 1.0264. (2)$\mu$ is between 0.7058 and 1.0072. We can see that there is a difference between the intervals.
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{0.751+0.841+...+0.881}{100}=0.8565.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(0.751-0.8565)^2+...+(0.881-0.8565)^2}{99}}=0.0518.$ $\alpha=1-0.95=0.05.$ $\sigma$ is unknown, hence we use the t-distribution with $df=sample \ size-1=100-1=99$ in the table. $t_{\alpha/2}=t_{0.025}=1.984.$ Margin of error:$t_{\alpha/2}\cdot\frac{s}{\sqrt {n}}=1.984\cdot\frac{0.8565}{\sqrt{100}}\approx0.1699.$ Hence the confidence interval:$\mu$ is between 0.8565-0.1699=0.6866 and 0.8565+0.1699=1.0264. Now using the given assumption: Margin of error:$t_{\alpha/2}\cdot\frac{s}{\sqrt {n}}\cdot \sqrt{\frac{N-n}{N-1}}=1.984\cdot\frac{0.8565}{\sqrt{100}}\cdot \sqrt{\frac{465-100}{465-1}}\approx0.1507.$ Hence the confidence interval:$\mu$ is between 0.8565-0.1507=0.7058 and 0.8565+0.1507=1.0072. We can see that there is a difference between the intervals.