Answer:
reject H0
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.
![H_0:\bar x =1.85\\H_a: \bar x \geq 1.85](https://img.qammunity.org/2020/formulas/mathematics/college/vifn5wlo7jjxe7j4lel0dczqed5xi4hm7p.png)
(right tailed test at 5% level of significance)
![\bar x = 1.70\\s = 0.4\\Se = (0.4)/(√(35) ) \\=0.0676](https://img.qammunity.org/2020/formulas/mathematics/college/xa5jxlh42s3fq6v8vzfswyxqc0arefaqcg.png)
Test statistic t = mean difference/std error =
![(1.70-1.85)/(0.0676) \\=-2.22](https://img.qammunity.org/2020/formulas/mathematics/college/mubsp7s56dd7exauag6i4zc8a1t5iy5k60.png)
df = 34
t critical = -1.697
Since our test statistic < t critical we reject null hypothesis..