Question

Nationally, about 11% of the total U.S. wheat crop is destroyed each year by hail.† An insurance company is studying wheat hail damage claims in a county in Colorado. A random sample of 16 claims in the county reported the percentage of their wheat lost to hail. 13 6 10 13 12 20 16 9 7 8 25 18 15 9 14 6

The sample mean is x = 12.6%. Let x be a random variable that represents the percentage of wheat crop in that county lost to hail. Assume that x has a normal distribution and σ = 5.0%. Do these data indicate that the percentage of wheat crop lost to hail in that county is different (either way) from the national mean of 11%? Use α = 0.01.

(a) What is the level of significance?

State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H₀: μ = 11%; H₁: μ ≠ 11%; two-tailed

H₀: μ = 11%; H₁: μ > 11%; right-tailed  

  H₀: μ ≠ 11%; H₁: μ = 11%; two-tailed

H₀: μ = 11%; H₁: μ < 11%; left-tailed

(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.

The standard normal, since we assume that x has a normal distribution with known σ.

The Student's t, since we assume that x has a normal distribution with known σ.    

The standard normal, since we assume that x has a normal distribution with unknown σ.

The Student's t, since n is large with unknown σ.

What is the value of the sample test statistic? (Round your answer to two decimal places.)


(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)

Answer 1

Givem that a sample of 16 is taken into consideration hence, n= 16

and Population mean , and the sample mean in %age

also given that the standard deviation as also given

Hence according to question

a) Level of significance as 10% or 0.01 .

The Hypotheses are

Hence two tail test is aaplied.

b) For this Experiment we use The standard normal distributiion since we assume that X has a normal disribution with known .

So, Test Statistics used is Z Hence Z can be calculated as

Hence Z calculated as Z=1.28

c) P value compued from Z statistics table or can be calculated on Calculator Hence P value as

P value-0.2005