Mathematics, 07.03.2020 04:12 hibbo579

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.

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Answer from: bryanmcmillianjr

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 from: tae8002001

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 .

Answer from: Quest

b. the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

Answer from: Quest

8 2/3 is the answer

step-by-step explanation: just use a calculator

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