It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each day. A doctor is concerned that her pregnant patients are not getting enough vitamin C. So, she collects data on 45 of her patients and finds that the mean vitamin intake of these 45 patients is 81 milligrams per day with a standard deviation of 12 milligrams per day. Based on a level of significance of α = .02, test the hypothesis.H0 : u = 85H1 : u< 85n = 45sample mean = 81σ = 12z based on α = .02 = -2.05 ( if lessthan -2.05 we will reject the null hypothesis)Sample standard deviation = 12/√45 = 1.789z= x-u/σ(o sample) - 81-85/1.789 = -2.236-2.236is less than -2.05 so wereject the H0 : u = 85

Respuesta :

Answer:

|Z| = 2.237 < 2.32 at 0.02 level of significance

Null hypothesis is accepted

Recommended that pregnant women not over eighteen years old get 85 milligrams of vitamin C each day

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 45

Mean of the Population 'μ' = 85

Mean of the sample 'x⁻' = 81

Standard deviation of the sample 'σ' = 12 milligrams

Null Hypothesis:   H₀:μ=85

Alternative Hypothesis:   H₁:μ≠ 85

Step(ii):-

Test statistic

                     [tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]

                    [tex]Z = \frac{81 -85}{\frac{12}{\sqrt{45} } }[/tex]

                  Z = -2.237

Level of significance = 0.02

Z₀.₀₂ = 2.32

|Z| = 2.237 < 2.32 at 0.02 level of significance

Conclusion:-

Null hypothesis is accepted

Recommended that pregnant women not over eighteen years old get 85 milligrams of vitamin C each day