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Answer:
a) ii. This is a left-tailed test.
b) -1.59
c) -1.301
d) i. reject null hypothesis
e) Option i) The data supports the claim that college students get less sleep than the general population.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 7 hours
Sample mean, = 6.87 hours
Sample size, n = 45
Alpha, α = 0.10
Sample standard deviation, s = 0.55 hours
First, we design the null and the alternate hypothesis
a) We use one-tailed(left) t test to perform this hypothesis.
b) Formula:
Putting all the values, we have
c) Now,
Since,
d) We fail to accept the null hypothesis and reject it.
We accept the alternate hypothesis and conclude that mean number of hours of sleep for all college students is less than 7 hours.
e) Option i) The data supports the claim that college students get less sleep than the general population.
Here is how I completed the problem:
I started by creating a ratio:
<em>Where 'x' is the desired amount of the 18% solution of sulfuric acid.
</em>From here, I solved for 'x':
<em>
</em>
∴You will need to add 1080ml of 18% Sulfuric Acid in order to obtain a 15% solution.
I started by creating a ratio:
<em>Where 'x' is the desired amount of the 18% solution of sulfuric acid.
</em>From here, I solved for 'x':
<em>
</em>
∴You will need to add 1080ml of 18% Sulfuric Acid in order to obtain a 15% solution.