Question

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. The director would like to test the hypothesis that H0:μ=2000 vs. HA:μ>2000. In a random sample of 50 students the director found that the average was 2105 calories/day with a standard deviation of 288 calories/day. Calculate the appropriate test statistic for this situation. Round your answer to 3 decimal places.

Answer 1

Test statistics = 2.578

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,
`H_0` :
`\mu` = 2000

Alternate Hypothesis,
`H_1` :
`\mu` > 2000

The test statistics used here will be;

T.S. =
`(Xbar -\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where, X bar = sample mean = 2105 calories/day

s = sample standard deviation = 288 calories/day

n = sample of students = 50

So, test statistics =
`(2105 -2000)/((288)/(√(50) ) )` ~
`t_4_9`

= 2.578

Therefore, the appropriate test statistic for this situation is 2.578 .

Answer 2

The calculated test statistic is 2.578

Explanation:

We are given the following in the question:

Population mean, μ = 2,000 calories/day

Sample mean,
`\bar{x}` = 2105

Sample size, n = 50

Alpha, α = 0.05

Sample standard deviation, s = 288

First, we design the null and the alternate hypothesis

`H_(0): \mu = 2000\text{ calories per day}\\H_A: \mu > 2000\text{ calories per day}`

We use one-tailed t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(2105 - 2000)/((288)/(√(50)) ) = 2.578`

Thus, the calculated test statistic is 2.578