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the mean of this sample is 264.4 yards. his driving distance is known to be normally distributed with a population standard deviation of 20 yards. test jean's claim that jack's driving distance is less than 275 yards at a 0.05 significance level. calculate the test statistic to 2 decimal places.

Respuesta :

After solving, the test statistic value of z is -2.65.

In the given question,

Sample Mean x=264.4

Population standard deviation σ=20

sample size (n) = 25

significance level α=0.05

To test  Jean's claim that Jack's driving distance is less than 275 yards

The null and alternative hypothesis are

H(0): μ=275

H(α): μ<275

Test Statistic Z is:

Z = (x-μ)/(σ/√n)

Z = (264.4-275)/(20/√25)

Z = -10.6/(20/5)

Z = -10.6/4

Z = -2.65

Hence, the test statistic value of z is -2.65.

To learn more about test statistic z link is here

brainly.com/question/3006526

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The right question is:

Jack tells Jean that his average drive of a golf ball is 275 yards. Jean is skeptical, she thinks he has overstated is driving distance and asks for substantiation. So Jack takes a random sample of 25 of his drives. The mean of this sample is 264.4 yards. His driving distance is known to be normally distributed with a population standard deviation of 20 yards. Test Jean's claim that Jack's driving distance is less than 275 yards at a 0.05 significance level. Calculate the test statistic to 2 decimal places.

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