Answer:
reject H0
Step-by-step explanation:
Given that a researcher is testing the claim that adults consume an average of at least 1.85 cups of coffee per day.
A sample of 35 adults shows a sample mean of 1.70 cups per day with a sample standard deviation of 0.4 cups per day.
[tex]H_0:\bar x =1.85\\H_a: \bar x \geq 1.85[/tex]
(right tailed test at 5% level of significance)
[tex]\bar x = 1.70\\s = 0.4\\Se = \frac{0.4}{\sqrt{35} } \\=0.0676[/tex]
Test statistic t = mean difference/std error =[tex]\frac{1.70-1.85}{0.0676} \\=-2.22[/tex]
df = 34
t critical = -1.697
Since our test statistic < t critical we reject null hypothesis..
