Answer: No
Step-by-step explanation:
Let [tex]\mu[/tex] be the population mean .
As per given , we have to test the hypothesis.
[tex]H_0:\mu=28.5\\\\ H_a:\mu>28.5[/tex]
∵ Alternative hypothesis ([tex]H_a[/tex]) is right-tailed , so our test is a right-tailed test.
Also, the population standard deviation is unknown to be 0.8 , so we use t-test.
Test statistic:[tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
, where [tex]\overline{x}[/tex] = Sample mean
[tex]\mu[/tex] = population mean
[tex]s[/tex] = sample standard deviation.
n= Sample size
Substitute [tex]\overline{x}=27.8[/tex]
[tex]s=4.1[/tex]
n= 100 , we get
[tex]t=\dfrac{27.8-28.5}{\dfrac{4.1}{\sqrt{100}}}[/tex]
[tex]t=\dfrac{-0.7}{\dfrac{4.1}{\sqrt{10}}}\approx-1.71[/tex]
By t-distribution, the critical t-value for degree of freedom 99 ( df =n-1) and significance level 0.10 :
[tex]t_{\alpha,df}=1.29[/tex]
Decision : ∵ Calculated -value (-1.71) < Critical value (1.29).
It means we do not reject the null hypothesis.
Conclusion : We do not have sufficient evidence at α = 0.10 significance level to reject the claim that the mean consumption of bottled water by a person in the United States is 28.5 gallons per year