Answer:
At 90% confidence interval, the estimate of the mean breaking weight is (770.45, 778.15)
Step-by-step explanation:
Given that:
sample size n =43
sample mean x = 774.3
standard deviation = 15.4
confidence interval = 90%
At C.I of 90% , the level of significance ∝ = 1 - C.I
the level of significance ∝ = 1 - 0.90
the level of significance ∝ = 0.10
The critical value for z at this level of significance is [tex]z_{\alpha/2} = z_{0.10/2}[/tex]
[tex]z_{0.05}[/tex] = 1.64
The margin of error can be computed as follows:
Margin of error = [tex]\mathtt{z_{\alpha/2} \times \dfrac{\sigma}{\sqrt{n}}}[/tex]
Margin of error = [tex]\mathtt{1.64 \times \dfrac{15.4}{\sqrt{43}}}[/tex]
Margin of error = [tex]\mathtt{1.64 \times \dfrac{15.4}{6.5574}}[/tex]
Margin of error = [tex]\mathtt{1.64 \times2.3485}[/tex]
Margin of error = 3.8515
The mean breaking weight for the 90% confidence interval is = [tex]\mathtt{\overline x \pm E < \mu }[/tex]
= [tex]\mathtt{\overline x - E < \mu < \overline x + E}[/tex]
= ( 774.3 - 3.8515 < μ < 774.3 + 3.8515 )
= (770.4485, 778.1515)
[tex]\simeq[/tex] (770.45, 778.15)
