Nationally, about 13.5% of the total U.S. wheat crop is destroyed each year by h
ID: 3231870 • Letter: N
Question
Nationally, about 13.5% of the total U.S. wheat crop is destroyed each year by hail. An insurance company is studying wheat hail damage claims in Weld County, Colorado. A random sample of 16 claims in Weld County reported the following percentages of their wheat lost to hail.
15, 8, 9, 11, 12, 20, 14, 11, 7, 10, 24, 20, 13, 9, 12, 5
Let x be a random variable that represents the percentage of wheat crop in Weld County lost to hail. Assume that x has a normal distribution and = 5.8%. Do these data indicate that the percentage of wheat crop lost to hail in Weld County is different (either way) from the national mean of 13.5%? Use = 0.05.
(a) Enter the following.
x = ____ %
s = ___ %
(b) Identify the claim, the null hypothesis, and the alternative hypothesis.
(c) Will you use a left-tailed, right-tailed, or two-tailed test?
left-tailed
right-tailed
two-tailed
(d) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.
The standard normal, since n is large with unknown . The standard normal, since n is large with known .
The Student's t, since we assume that x has a normal distribution with known .
The standard normal, since we assume that x has a normal distribution with known .
The Student's t, since n is large with unknown .
The standard normal, since we assume that x has a normal distribution with unknown .
(e) Sketch the sampling distribution showing the area corresponding to the approximate P-value.
(f) Will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?
At the = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.
At the = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
At the = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
(g) State your conclusion in the context of the application.
Fail to reject the null hypothesis, there is insufficient evidence that the average hail damage to wheat crops in Weld County differs from the national average.
Fail to reject the null hypothesis, there is sufficient evidence that the average hail damage to wheat crops in Weld County differs from the national average.
Reject the null hypothesis, there is insufficient evidence that the average hail damage to wheat crops in Weld County differs from the national average.
Reject the null hypothesis, there is sufficient evidence that the average hail damage to wheat crops in Weld County differs from the national average.
Claim: < > = 13.5% Ho: < > = 13.5% H1: < > = 13.5%Explanation / Answer
Given that,
population mean(u)=13.5
standard deviation, =5.8
sample mean, x =12.5
number (n)=16
null, Ho: =13.5
alternate, H1: !=13.5
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 12.5-13.5/(5.8/sqrt(16)
zo = -0.68966
| zo | = 0.68966
critical value
the value of |z | at los 5% is 1.96
we got |zo| =0.68966 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.68966 ) = 0.49041
hence value of p0.05 < 0.49041, here we do not reject Ho
ANSWERS
---------------
null, Ho: =13.5
alternate, H1: !=13.5
test statistic: -0.68966
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.49041
x=12.5
s=5.1381%
two-tailed
The standard normal, since we assume that x has a normal distribution with known
At the = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant
Fail to reject the null hypothesis, there is insufficient evidence that the average hail damage to wheat crops in Weld County differs from the national average.
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