Based on the sample of n = 25 healthy-weight students, can you conclude that healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population? Use a one-tailed test with LaTeX: \alphaα. = .05

Step-by-step explanation:

Null hypothesis, H₀ : μ ≥ 4.22

Alternative hypothesis, H₁ : μ < 4.22

Level of significance, ∝ = 0.05

Test statistic value Z = M₂ - μ / (δ/ √n)

                                   = 4.01 - 4.22 / 0.60 / √25

                                    = -1.75

P- value

P-value = P(Z < -1.75)

= P( Z ≤ -1.75)

= (=NORMSDIST(1.75))             (Use Ms Excel Function)

= 0.04005916

=0.0401

Decision: Reject  H₀ because the P-value is less than the 0.05 level of significance.

Conclusion, there is sufficient evidence that the number of snacks eaten by healthy-weight students is significantly less than the number for the general population at the 0.05 level of significance.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.

rounding of = 42%

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