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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.

Answers

Answer 1

Answer:

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

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

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

Answer 2

Answer:

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. = $(Xbar -\mu)/((s)/(√(n) ) )$ ~ $t_n_-_1$

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 .

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