Answer:
z = -1.645
Step-by-step 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.
Answer:
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.
