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

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, $\bar{x}$ = 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:

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

Putting all the values, we have

$t_{stat} = \displaystyle\frac{2105 - 2000}{\frac{288}{\sqrt{50}} } = 2.578$

Thus, the calculated test statistic is 2.578

Answer 2

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, $H_0$ : $\mu$ = 2000  

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

The test statistics used here will be;

                 T.S. = $\frac{Xbar -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

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

              s = sample standard deviation = 288 calories/day

              n = sample of students = 50

So, test statistics = $\frac{2105 -2000}{\frac{288}{\sqrt{50} } }$ ~ $t_4_9$

                            = 2.578

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