Hi, can you help me answer this question please, thank you

Step 1: Find the t-value
[tex]\begin{gathered} \text{First, subtract 100\% by the confidence level},\text{ and divide it by two} \\ \frac{100\%-95\%}{2}=\frac{5\%}{2}=2.5\% \\ \\ \text{Converted to decimal that is} \\ 2.5\%\rightarrow0.025 \\ \text{Next find the degree of freedom }df \\ df=n-1 \\ df=19-1 \\ df=18 \\ \\ \text{Now that we have degree of freedom, find in the t-distribution table our t-value} \\ df=18,\text{ and }0.025 \end{gathered}[/tex]The intersection is 2.101, which means that our t-value is 2.101.
Step 2: Calculate the standard error
[tex]\begin{gathered} \text{SE}=\frac{s}{\sqrt[]{n}}=\frac{4}{\sqrt[]{19}} \\ \text{SE}=0.9176629355 \end{gathered}[/tex]Step 3: Multiply the t-value to the standard error
[tex]\begin{gathered} t\times SE=2.101\times0.9176629355 \\ =1.928009827\rightarrow1.928 \end{gathered}[/tex]Step 4: Add and Subtract it to the sample mean, to get the lower and upper limit.
[tex]\begin{gathered} \text{Upper limit} \\ \bar{x}+1.928 \\ 46+1.928=47.928 \\ \\ \text{Lower limit} \\ \bar{x}-1.928 \\ 46-1.928=44.072 \end{gathered}[/tex]Final Step: Now that we have Upper and Lower limit, our confidence interval is:
[tex]44.072<\mu<47.928[/tex]