Question

6112.2673580.qx3zay?Jump to level 1The additional growth of plants in one week are recorded for 6 plants with asample standard deviation of 1 inches and sample mean of 8 inches.t* at the 0.05 significance level = Ex: 1.234Margin of error = Ex: 1.234How was this section?Confidence interval = [Ex: 12.345 Ex: 12.345][smaller value, larger value]CheckNextProvide feedback2Fe

Answer 1

Given the confidence interval formula as shown below.

`CI=\bar{x}\pm t_{(\alpha)/(2)}*(s)/(√(n))`
`\begin{gathered} \bar{x}\rightarrow mean \\ s\rightarrow standard\text{ deviation} \\ n\rightarrow sample\text{ size} \\ t_{(\alpha)/(2)}\rightarrow critical\text{ value} \end{gathered}`

The margin of error is

`MOE=\pm t_{(\alpha)/(2)}*(s)/(√(n))`

Using an online calculator for the critical value

`t_{(\alpha)/(2)}\text{ at 0.05 significance level is}\rightarrow\pm2.571`

The Margin of error MOE is given as:

`\begin{gathered} MOE=\pm t_{(\alpha)/(2)}*(s)/(√(n)) \\ \pm t_{(\alpha)/(2)}=\pm2.5706 \\ s=1,n=6 \\ (s)/(√(n))=(1)/(√(6))=0.4082 \\ MOE=\pm2.5706*0.4082=\pm1.049 \end{gathered}`

Thus, the confidence interval is:

`CI=\bar{x}\text{ }\pm MOE`
`\begin{gathered} \bar{x}=8,MOE=\pm1.049 \\ CI=8\pm1.049 \\ CI\rightarrow(6.951,9.049) \end{gathered}`