The Solution:
Given:
[tex]\begin{gathered} \bar{X}=734.2\text{ grams} \\ N=200 \\ \sigma=15.7\text{ grams} \\ \bar{x}=? \end{gathered}[/tex]Using a 10% level of significance, we have
[tex]Z_{\alpha}=Z_{0.1}=P(x<-Z\cup x>Z)=1.645_{}[/tex][tex]\begin{gathered} 1.645=\frac{\sqrt[]{200}(\mu-734.2)}{15.7} \\ \text{ Cross multiplying, we get} \\ \sqrt[]{200}(\mu-734.2)=1.645\times15.7=25.8265 \end{gathered}[/tex]So,
[tex]\begin{gathered} \mu-734.2=\frac{25.8265}{\sqrt[]{200}} \\ \\ \mu=734.2+1.82621=736.026 \end{gathered}[/tex]Therefore, comparing the population mean with the sample mean, we can see that there is not much difference between them. Hence, we fail to reject the null hypothesis.
