Question

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

Answer 1

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 2

z = -1.645

Explanation:

b) n =25, M = 4.01 , \mu = 4.22 , \sigma = 0.60 , \alpha= 0.05

The hypothesis are given by,

H0 : \mu\geq 4.22 v/s H1 : \mu < 4.22

The test statistic is given by,

Check attachment for the formula that should br here.

z = \frac{4.01- 4.22 }{0.60 /\sqrt{25}}

= -1.75

The critical value of z = -1.645

The calculated value z > The critical value of z

Hence we reject null hypothesis.

The healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population.