It is claimed that an automobile is driven on the average more than 20,000 kilometers per year. To this claim, a random sample of 20 automobile owners is asked to keep a record of the kilometers they travel. Would you agree with the claim if the random sample showed an average of 23,500 kilometers and standard deviation of 3900 kilometers? Use alpha = 5%.

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Step-by-step explanation:

Given the following :

Sample mean(x) = 23500

Sample standard deviation (sd) = 3900

Sample size (n) = 20

Population mean (m) = 20,000

Null hypothesis : m = 20000

H1: m > 20000

To obtain the z-score :

(population mean - sample mean) / (sample standard deviation /√sample size)

(x - m) / (sd/√n)

(23500 - 20000) / (3900 / √20)

3500 / (3900 /4.4721359)

3500 / 872.06651

= 4.0134

Get the P value to know if to reject or accept the null:

P(z > 4.0134) = 1 - P(z < 4.0134)

P(z < 4.0134) = 1

1 - P(z < 4.0134) = 1 - 1 =0

Since P value is < alpha, we reject the null.

Hence average is > 20000

5.25 ounces

step-by-step explanation:

divided 3/4 and 6

answer: a) 56% women

b)44% men

step-by-step explanation: