Let [tex]\mu[/tex] be the population mean .
By considering the given information in the question , we have
[tex]\text{Null hypothesis }H_0: \mu=20\\\\\text{Alternative hypothesis } H_a: \mu\neq20[/tex]
Here, [tex]H_a[/tex] is two-tailed , so the test is a two-tailed test.
Also, population standard deviation is unknown , so we perform a two-tailed t-test.
For sample size : n= 18
Sample mean : [tex]\overline{x}=20.45[/tex]
Sample standard deviation : s=0.80
Test statistic :
[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t=\dfrac{20.45-20}{\dfrac{0.80}{\sqrt{18}}}[/tex]
[tex]t=\dfrac{0.45}{\dfrac{0.80}{4.2426}}[/tex]
[tex]t=\dfrac{0.45}{0.1885636}[/tex]
[tex]t\approx 2.39[/tex]
Two-tailed Critical values for [tex]\alpha=0.05[/tex] and degree of freedom df=n-1=17
[tex]t_{(\alpha/2,\ df)}=t_{(0.025,\ 17)}=\pm2.11[/tex]
Decision : Since the calculated t-value(2.39) does not lie between the critical values -2.11 and 2.11.
So we reject the null hypothesis .
Conclusion : We have enough evidence at 0.05 significance level to support the claim that the sample estimate is different from the company’s expected weight of 20 oz.
