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atroni [7]
3 years ago
15

A simple random sample of 60 adults is obtained from a normally distributed​ population, and each​ person's red blood cell count

​ (in cells per​ microliter) is measured. The sample mean is 5.25 and the sample standard deviation is 0.55. Use a 0.05 significance level and the given calculator display to test the claim that the sample is from a population with a mean less than 5.4 comma which is a value often used for the upper limit of the range of normal values. What do the results suggest about the sample​ group?
Mathematics
1 answer:
Goshia [24]3 years ago
7 0

Answer:

Step-by-step explanation:

sample size = n = 60

Sample mean = x bar = 5.25

Sample sd = s = 0.55

Std error of sample = s/sqrt n = 0.071

H_0: x bar = 5.4\\H_a: x bar

(One tailed test)

Mean difference = 5.25-5.4 = -0.15

Test statistic t = Mean diff/Se = -2.11

df=59

(since population std dev is not known t test is used)

p value =0.01955

Since p <alpha =0.05 we reject null hypothesis.

The results suggest about the sample​ group that the sample is from a population with a mean less than 5.4

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6 0
2 years ago
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Answer:

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Step-by-step explanation:

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from his current situation we can say that ,

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so  for working of 1 hour , Maddox monthly income wages = $1760÷40=$44

hence for working of 25 hour , Maddox monthly income wages will be $44 × 25 = $1100.

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At last lets see Net income of modified budget A. he is able to save $30 , hence we can say that best fit for his new situation is modified budget A


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