Velvetleaf is a particularly annoying weed in corn fields. It produces lots of s
ID: 3339387 • Letter: V
Question
Velvetleaf is a particularly annoying weed in corn fields. It produces lots of seeds, and the seeds wait in the soil for years until conditions are right. The velvetleaf seed beetle feeds on the seeds and might be a natural weed control. Here are the total seeds, seeds infected by the beetle, and percent of seeds infected for 28 velvetleaf plants:
Seeds Infected Percent 2450 135 5.6 2504 101 3.8 2114 76 3.5 1110 24 2.4 2137 121 5.5 8015 189 2.6 1623 31 2.0 1531 44 2.7 2008 73 3.8 1716 12 0.6
Seeds Infected Percent 721 27 3.6 863 40 4.8 1136 41 3.8 2819 79 2.7 1911 82 4.1 2101 85 4.2 1051 42 4.2 218 0 0.0 1711 64 3.8 164 7 4.1
Seeds Infected Percent 2228 156 6.8 363 31 8.4 5973 240 4.1 1050 91 8.9 1961 137 7.0 1809 92 5.1 130 5 4.0 880 23 2.4
In R, do a complete analysis of the percent of seeds infected by the beetle. #R Code
Percent = c( 5.6, 3.8, 3.5, 2.4, 5.5, 2.6, 2.0, 2.7, 3.8, 0.6, 3.6, 4.8, 3.8, 2.7, 4.1, 4.2, 4.2, 0.0, 3.8, 4.1, 6.8, 8.4, 4.1, 8.9, 7.0, 5.1, 4.0, 2.4)
hist(Percent)
mean(Percent)
sd(Percent)
t.test(Percent, mu = 5.0, conf.level = 0.90)
Researchers would like to test whether the average percent of seeds infected is different than 5 percent. What is the null and alternative hypothesis for this test?
H0: = 5 percent Ha: 5 percent
H0: p = 5 percent Ha:p 5 percent
H0: < 5 percent Ha: = 5 percent
H0: p 5 percent Ha: p = 5 percent
H0: > 5 percent Ha: < 5 percent
What is the average and standard deviation for the percent of seeds that are infected? (Round your answers to four decimal places.)
average x =
s =
According to the R output, the test statistic and p-value are: (Round your answers to four decimal places.)
t =
p-value =
Given a significance level of 0.10. What can we conclude from the hypothesis test?
Fail to reject the null hypothesis.
Reject the null hypothesis.
Include a 90% confidence interval for the mean percent infected in the population of all velvetleaf plants. (Round your answers to two decimal places.)
( )% to ( )%
Do you think that the beetle is very helpful in controlling the weed?
The beetle infects less than 5% of seeds, so it is unlikely to be effective in controlling velvetweed.
The beetle infects more than 5% of seeds, so it is likely to be effective in controlling velvetweed.
The beetle infects more than 15% of seeds, so it is likely to be effective in controlling velvetweed.
The beetle infects less than 15% of seeds, so it is unlikely to be effective in controlling velvetweed. Show My Work (Optional)
Show Your Work
Explanation / Answer
1)Researchers would like to test whether the average percent of seeds infected is different than 5 percent. What is the null and alternative hypothesis for this test?
Ans - H0: = 5 percent Ha: 5 percent
Because average percent of seeds infected is different than 5 percent that means it either less or greter that's why we use not equal to sign.
What is the average and standard deviation for the percent of seeds that are infected?
average x = 4.0892
> mean(Percent)
[1] 4.089286
s = 2.0124
> sd(Percent)
[1] 2.012432
According to the R output, the test statistic and p-value are: (Round your answers to four decimal places.)
> t.test(Percent, mu = 5.0, conf.level = 0.90)
One Sample t-test
data: Percent
t = -2.3946, df = 27, p-value = 0.02384
alternative hypothesis: true mean is not equal to 5
90 percent confidence interval:
3.441502 4.737070
sample estimates:
mean of x
4.089286
t = -2.3946
p-value = 0.02384
Given a significance level of 0.10. What can we conclude from the hypothesis test?
Reject the null hypothesis.
Because P-value is less than 0.10 that's why we Reject the null hypothesis .
Include a 90% confidence interval for the mean percent infected in the population of all velvetleaf plants. (Round your answers to two decimal places.)
90 percent confidence interval : ( 3.4415 , 4.7370 )
Do you think that the beetle is very helpful in controlling the weed?
The beetle infects less than 5% of seeds, so it is unlikely to be effective in controlling velvetweed.
Beucuse P-value is less than 0.05 thats why Answer is this .
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Best of luck :)
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