Respuesta :
Answer:
The calculated test statistic is 2.578
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 2,000 calories/day
Sample mean, [tex]\bar{x}[/tex] = 2105
Sample size, n = 50
Alpha, α = 0.05
Sample standard deviation, s = 288
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 2000\text{ calories per day}\\H_A: \mu > 2000\text{ calories per day}[/tex]
We use one-tailed t test to perform this hypothesis.
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{2105 - 2000}{\frac{288}{\sqrt{50}} } = 2.578[/tex]
Thus, the calculated test statistic is 2.578
Answer:
Test statistics = 2.578
Step-by-step explanation:
We are given that the director of student health at a large university was concerned that students at his school were consuming too many calories each day. For a certain population of college-age students, it is recommended to consume around 2,000 calories/day.
Also, given Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 2000
Alternate Hypothesis, [tex]H_1[/tex] : [tex]\mu[/tex] > 2000
The test statistics used here will be;
T.S. = [tex]\frac{Xbar -\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, X bar = sample mean = 2105 calories/day
s = sample standard deviation = 288 calories/day
n = sample of students = 50
So, test statistics = [tex]\frac{2105 -2000}{\frac{288}{\sqrt{50} } }[/tex] ~ [tex]t_4_9[/tex]
= 2.578
Therefore, the appropriate test statistic for this situation is 2.578 .
