Answer:
The confidence interval will be "41.04, 43.81".
Explanation:
Given that,
Sample size,
n = 10
Sample total,
423.5
Sample mean,
![\bar X =(423.5)/(10)](https://img.qammunity.org/2022/formulas/mathematics/college/v4wgh63ojt5xhhdb1pwfn6e8qf29sno1ws.png)
Sample variance,
![s = √(4.5894)](https://img.qammunity.org/2022/formulas/mathematics/college/aw6dnr87rj729wl9r95soxofhuq46bwxdv.png)
![=2.1423](https://img.qammunity.org/2022/formulas/mathematics/college/mal4wxa9zx5v3r1it0xzc4xugijfm779je.png)
![df=n-1](https://img.qammunity.org/2022/formulas/mathematics/college/fmtaxsgm6s5btzynowjmhhm6xrd73sttdr.png)
![=9](https://img.qammunity.org/2022/formulas/mathematics/high-school/5jjdy5rmda5ssfu2fmg0extuyfqqd9gn7s.png)
![t^*=18331](https://img.qammunity.org/2022/formulas/mathematics/college/xndjs7294wa336v5ucjgmsb1d3fktpk83p.png)
Now,
The margin of error will be:
⇒
![E=(s* t^*)/(√(n) )](https://img.qammunity.org/2022/formulas/mathematics/college/wznq075twbj65m6ppqhngscv3cntcnwy3x.png)
![=(2.1423* 1.8331)/(√(9) )](https://img.qammunity.org/2022/formulas/mathematics/college/23pcoauct79j849fen1u4xxkk56hk8tn0v.png)
![=(3.928)/(3 )](https://img.qammunity.org/2022/formulas/mathematics/college/wqkqut3h3mxauhualpwc7apeps8o48e25t.png)
![=1.309](https://img.qammunity.org/2022/formulas/mathematics/college/6dxmtupp9g7j2ajcp1fsf2xn4ztecl8ngc.png)
hence,
The 90% confidence level will be:
=
![(\bar X-E),(\bar X+E)](https://img.qammunity.org/2022/formulas/mathematics/college/oyvsgm66v24gvaztnmh4z5j4stvouqxo7b.png)
By substituting the values, we get
=
![(42.35-1.309),(42.35+1.309)](https://img.qammunity.org/2022/formulas/mathematics/college/dft8ubh8wsbktgeoy09cxetrlkc6patdti.png)
=
![(41.04),(43.81)](https://img.qammunity.org/2022/formulas/mathematics/college/hch0j10earvil19w7eronevprfhzk5zhcx.png)