We will answer only the first question in the first picture - as per policy.
We have a normally distributed population, and we have the following information:
• The confidence level is C = 0.95
,
• The sample mean is x-bar = 12.3
,
• The sample standard deviation is s = 3.0
,
• The sample size, n = 8
And we have to find the confidence interval for the population mean, μ, using the t-distribution.
To find it, we can proceed as follows:
1. We have that the confidence interval for the population's mean, μ, using the t-distribution is given by:
[tex]\bar{x}\pm t^{\ast}(\frac{s}{\sqrt{n}})[/tex]
And we already know that the sample size is less than 30 (n < 30), the population's standard deviation is unknown, and the population is normally distributed.
2. Now, we have to find the value for t^* as follows:
[tex]\begin{gathered} t^{\ast}=\frac{1-0.95}{2}=0.025 \\ \\ t^{\ast}=0.025 \end{gathered}[/tex]
3. Now, we need to find the critical value for the t-distribution using the inverse cumulative function using the degrees of freedom, n - 1. In this case, n = 8. Then the degree of freedom is 7. Then the inverse cumulative function is:
[tex]\begin{gathered} \text{ invT\lparen0.025,7\rparen=}2.36462425159 \\ \\ t^{\ast}=2.36462425159\approx2.36 \end{gathered}[/tex]
4. Now, since the t-distribution is symmetrical, then we can use the formula for the confidence interval as follows:
[tex]\begin{gathered} \bar{x}\pm t^{\operatorname{\ast}}(\frac{s}{\sqrt{n}}) \\ \\ 12.3\pm2.36(\frac{3.0}{\sqrt{8}}) \end{gathered}[/tex]
5. Finally, the confidence interval is:
[tex]\begin{gathered} 12.3+2.5031580054=14.8031580054\approx14.8 \\ \\ 12.3-2.5031580054=9.7968419946\approx9.8 \\ \\ (9.8,14.8) \end{gathered}[/tex]
Therefore, in summary, the confidence interval is (9.8, 14.8).
