First we have to make 2 1/6 into an improper fraction. This can be rewritten as 13/6. Now, we can multiply 13 by 5 and get 65/6. If we re-convert this into a mixed number, 6 goes into 65 10 times with a remainder of 5, so the final answer is 10 5/6.

4 0

3 years ago

Assume the general population gets an average of 7 hours of sleep per night. You randomly select 45 college students and survey

tekilochka [14]

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, $\bar{x}$ = 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

$H_{0}: \mu = 7\text{ hours}\\H_A: \mu < 7\text{ hours}$

a) We use one-tailed(left) t test to perform this hypothesis.

b) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{6.87 - 7}{\frac{0.55}{\sqrt{45}} } =-1.59$

c) Now,

$t_{critical} \text{ at 0.10 level of significance, 44 degree of freedom } = -1.301$

Since,

$t_{stat} < t_{critical}$

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.

8 0

3 years ago