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, [tex]\bar{x}[/tex] = 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
[tex]H_{0}: \mu = 7\text{ hours}\\H_A: \mu < 7\text{ hours}[/tex]
a) We use one-tailed(left) t test to perform this hypothesis.
b) Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{6.87 - 7}{\frac{0.55}{\sqrt{45}} } =-1.59[/tex]
c) Now,
[tex]t_{critical} \text{ at 0.10 level of significance, 44 degree of freedom } = -1.301[/tex]
Since,
[tex]t_{stat} < t_{critical}[/tex]
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.