In your work for a national health organization, you are asked to monitor the am
ID: 3230666 • Letter: I
Question
In your work for a national health organization, you are asked to monitor the amount of sodium in a certain brand of cereal. You find that a random sample of 52 cereal serving has a mean sodium content of 232 mg with a standard deviation of 10 mg. At a = 0.05, can you conclude that the mean sodium content per serving of cereal is no more than 230 mg? 43. H_0 = 44. H_a = 45. Suite the claim 46. a = 47. Critical value z-e = 48. Standardized Test Statistic 49. Decision The USDA reports the mean cost of raising a child from birth to age two in a rural in $8, 390. You believe that this value is incorrect, so you select a random sample of 900 children and find that the mean cost is $8275 with a standard deviation of $1, 540. a = 0.05 50. H_0 = 51. H-a = 52. State the claim 53. a = 54. Critical value Z_e = 55. Standardized Test Statistic 56. DecisionExplanation / Answer
43 - 49
Let X = amount of sodium (mg) in a certain cereal. We assume X ~ N(µ, 2).
Given sample size, n = 52, sample average, Xbar = 232 mg, sample standard deviation,
s = 10 mg
43) H0 : µ = µ0 = 230
44) H0 : µ > 230
45) Claim: mean amount of sodium in the particular cereal is no more than 230 mg
46) = 0.05
47) Critical value: tcrit = upper 5% point of tn – 1 = t51 = 1.676[using Excel Function]
48) Test statistic: t = (n)(Xbar - µ0)/s = (52)(232 - 230)/10 = 1.442
[Distribution: Under H0, t ~ tn – 1]
49) Decision: Since tcal < tcrit, H0 is accepted.
=> Statistical evidence is sufficient enough to validate the claim,
50 - 56
Let Y = cost ($) of raising a child from birth to age 2 in rural areas. We assume Y ~ N(µ, 2).
Given sample size, n = 900, sample average, Ybar = 8275, sample standard deviation,
s = 1540
50) H0 : µ = µ0 = 8390
51) H0 : µ 8390
52) Claim: USDA report of mean cost of raising a child from birth to age 2 in rural areas is $8390 is not correct.
53) = 0.05
54) Critical value: tcrit = upper 2.5% point of tn – 1 = t899 = 1.96 [using Excel Function]
55) Test statistic: t = (n)(Ybar - µ0)/s = (900)(8275 - 8390)/1540 = - 2.240
[Distribution: Under H0, t ~ tn – 1]
56) Decision: Since | tcal | > tcrit, H0 rejected.
=> Statistical evidence is sufficient enough to validate the claim,