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1 answer:
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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:
Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.
Step-by-step explanation:
We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.
So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;
Z = ~ N(0,1)
where, = average age of the random sample of horses with colic = 12 yrs
= average age of all horses seen at the veterinary clinic = 10 yrs
= standard deviation of all horses coming to the veterinary clinic = 8 yrs
n = sample of horses = 60
So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P(
12)
P(
12) = P(
) = P(Z
1.94) = 1 - P(Z < 1.94)
= 1 - 0.97381 = 0.0262
Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.