Question

The following data represent soil water content (percentage of water by volume) for independent random samples of soil taken from two experimental fields growing bell peppers. Soil water content from field I: x1; n1 = 72 15.3 11.3 10.2 10.8 16.6 8.3 9.1 12.3 9.1 14.3 10.7 16.1 10.2 15.2 8.9 9.5 9.6 11.3 14.0 11.3 15.6 11.2 13.8 9.0 8.4 8.2 12.0 13.9 11.6 16.0 9.6 11.4 8.4 8.0 14.1 10.9 13.2 13.8 14.6 10.2 11.5 13.1 14.7 12.5 10.2 11.8 11.0 12.7 10.3 10.8 11.0 12.6 10.8 9.6 11.5 10.6 11.7 10.1 9.7 9.7 11.2 9.8 10.3 11.9 9.7 11.3 10.4 12.0 11.0 10.7 8.9 11.2 Soil water content from field II: x2; n2 = 80 12.3 10.3 13.6 8.1 13.5 7.8 11.8 7.7 8.1 9.2 14.1 8.9 13.9 7.5 12.6 7.3 14.9 12.2 7.6 8.9 13.9 8.4 13.4 7.1 12.4 7.6 9.9 26.0 7.3 7.4 14.3 8.4 13.2 7.3 11.3 7.5 9.7 12.3 6.9 7.6 13.8 7.5 13.3 8.0 11.3 6.8 7.4 11.7 11.8 7.7 12.6 7.7 13.2 13.9 10.4 12.9 7.6 10.7 10.7 10.9 12.5 11.3 10.7 13.2 8.9 12.9 7.7 9.7 9.7 11.4 11.9 13.4 9.2 13.4 8.8 11.9 7.1 8.8 14.0 14.3 (a) Use a calculator with mean and standard deviation keys to calculate x1, s1, x2, and s2. (Round your answers to two decimal places.) x1 = s1 = x2 = s2 = (b) Let μ1 be the population mean for x1 and let μ2 be the population mean for x2. Find a 95% confidence interval for μ1 − μ2. (Round your answers to two decimal places.) lower limit upper limit (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? All negative? Of different signs? At the 95% level of confidence, is the population mean soil water content of the first field higher than that of the second field? Because the interval contains only positive numbers, we can say that the mean soil water content of the first field is higher. Because the interval contains both positive and negative numbers, we cannot say that the mean soil water content of the first field is higher. We can not make any conclusions using this confidence interval. Because the interval contains only negative numbers, we can say that the mean soil water content of the second field is higher. (d) Which distribution did you use (standard normal or Student's t)? Why? The Student's t-distribution was used because σ1 and σ2 are unknown. The Student's t-distribution was used because σ1 and σ2 are known. The standard normal distribution was used because σ1 and σ2 are unknown. The standard normal distribution was used because σ1 and σ2 are known. Do you need information about the soil water content distributions? Both samples are large, so information about the distributions is not needed. Both samples are small, so information about distributions is needed. (e) Use α = 0.01 to test the claim that the population mean soil water content of the first field is higher than that of the second. (i) What is the level of significance? State the null and alternate hypotheses. H0: μ1 = μ2; H1: μ1 ≠ μ2 H0: μ1 ≠ μ2; H1: μ1 = μ2 H0: μ1 = μ2; H1: μ1 > μ2 H0: μ1 = μ2; H1: μ1 < μ2 Correct: (ii) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate. (Test the difference μ1 − μ2. Do not use rounded values. Round your answer to three decimal places.) (iii) Find (or estimate) the P-value. P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005. Sketch the sampling distribution and show the area corresponding to the P-value. WebAssign Plot WebAssign Plot WebAssign Plot WebAssign Plot. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α? At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant. At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant. (v) Interpret your conclusion in the context of the application. Fail to reject the null hypothesis, there is sufficient evidence that the population mean soil water content of the first field is higher than that of the second. Reject the null hypothesis, there is insufficient evidence that the population mean soil water content of the first field is higher than that of the second. Reject the null hypothesis, there is sufficient evidence that the population mean soil water content of the first field is higher than that of the second. Fail to reject the null hypothesis, there is insufficient evidence that the population mean soil water content of the first field is higher than that of the sec

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=11.42
standard deviation , s.d1=2.07
number(n1)=72
y(mean)=10.661
standard deviation, s.d2 =3.009
number(n2)=80
b.
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((4.285/72)+(9.054/80))
= 0.416
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 71 d.f is 1.994
margin of error = 1.994 * 0.416
= 0.829
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (11.42-10.661) ± 0.829 ]
= [-0.07 , 1.588]
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DIRECT METHOD
given that,
mean(x)=11.42
standard deviation , s.d1=2.07
sample size, n1=72
y(mean)=10.661
standard deviation, s.d2 =3.009
sample size,n2 =80
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 11.42-10.661) ± t a/2 * sqrt((4.285/72)+(9.054/80)]
= [ (0.759) ± t a/2 * 0.416]
= [-0.07 , 1.588]
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c.
interpretations:
1. we are 95% sure that the interval [-0.07 , 1.588] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
d.
The Student's t-distribution was used because σ1 and σ2 are unknown.
e.
Given that,
mean(x)=11.42
standard deviation , s.d1=2.07
number(n1)=72
y(mean)=10.661
standard deviation, s.d2 =3.009
number(n2)=80
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.38
since our test is right-tailed
reject Ho, if to > 2.38
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =11.42-10.661/sqrt((4.2849/72)+(9.05408/80))
to =1.8265
| to | =1.8265
critical value
the value of |t α| with min (n1-1, n2-1) i.e 71 d.f is 2.38
we got |to| = 1.82646 & | t α | = 2.38
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.8265 ) = 0.03599
hence value of p0.01 < 0.03599,here we do not reject Ho
ANSWERS
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i.
level of significance =0.01
null, Ho: u1 = u2
alternate, H1: u1 > u2
ii.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. The Student's t.
test statistic: 1.8265
critical value: 2.38
decision: do not reject Ho
iii.
p-value: 0.03599
p value is greater than alpha value
iv.
At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.
v.
we do not have enough evidence to support the claim that the population mean soil water content of the first field is higher than that of the second.