Answer
$\mu$ is between 6.43 and 15.67. The population mean doesn't seem to be less than 7, because 7 is in the confidence interval.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{3+6.5+...+17.5}{10}=11.05.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(3-11.05)^2+...+(17.5-11.05)^2}{9}}=6.46.$
$\alpha=1-0.95=0.05.$ $\sigma$ is 6.46, hence we use the z-distribution with $df=sample \ size-1=10-1=9$ in the table. $z_{\alpha/2}=z_{0.025}=1.96.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=1.96\cdot\frac{6.46}{\sqrt{10}}=4.62.$ Hence the confidence interval:$\mu$ is between 11.05-4.62=6.43 and 11.05+4.62=15.67. The population mean doesn't seem to be less than 7, because 7 is in the confidence interval.