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 5.7	2504 101 3.9	2114 76 3.7	1110 24 2.4	2137 121 5.8	8015 189 2.3	1623 31 1.8	1531 44 2.9	2008 73 3.4	1716 12 0.9
Seeds Infected Percent	721 27 3.6	863 40 4.5	1136 41 3.8	2819 79 2.9	1911 82 4.3	2101 85 3.9	1051 42 3.9	218 0 0.0	1711 64 3.8	164 7 4.3
Seeds Infected Percent	2228 156 7.1	363 31 8.3	5973 240 4.2	1050 91 8.6	1961 137 7.1	1809 92 5.2	130 5 3.7	880 23 2.6

In R, do a complete analysis of the percent of seeds infected by the beetle.
#R Code

Percent = c(

5.7,

3.9,

3.7,

2.4,

5.8,

2.3,

1.8,

2.9,

3.4,

0.9,

3.6,

4.5,

3.8,

2.9,

4.3,

3.9,

3.9,

0.0,

3.8,

4.3,

7.1,

8.3,

4.2,

8.6,

7.1,

5.2,

3.7,

2.6)

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: μ = 5 percent

H_a: μ ≠ 5 percent

H₀: μ > 5 percent

H_a: μ < 5 percent

    

H₀: p = 5 percent

H_a:p ≠ 5 percent

H₀: μ < 5 percent

H_a: μ = 5 percent

H₀: p ≠ 5 percent

H_a: p = 5 percent

What is the average and standard deviation for the percent of seeds that are infected? (Round your answers to four decimal places.)

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?

Reject the null hypothesis.Fail to 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 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.The beetle infects less than 5% of seeds, so it is unlikely to be effective in controlling velvetweed.

• Show My Work

  (Optional)

Answer 1

The output of the R code is,

Percent = c(5.7,   3.9,   3.7,   2.4,   5.8,   2.3,   1.8,   2.9,   3.4,   0.9,
3.6,   4.5,   3.8,   2.9,   4.3,   3.9,   3.9,   0.0,   3.8,   4.3,   7.1,   8.3,   4.2,   8.6,
7.1,   5.2, 3.7,2.6)
length(Percent)
hist(Percent)
mean(Percent)

> mean(Percent)
[1] 4.092857

> sd(Percent)
[1] 1.997763

> t.test(Percent, mu = 5.0, conf.level = 0.90)

   One Sample t-test

data: Percent
t = -2.4028, df = 27, p-value = 0.02341
alternative hypothesis: true mean is not equal to 5
90 percent confidence interval:
3.449795 4.735920
sample estimates:
mean of x
4.092857

Null and Alternate hypotheses are,

H₀: μ = 5 percent

H_a: μ ≠ 5 percent

From the R output,

x	=4.0929
s	=1.9978

t = -2.4028

p-value = 0.0234

Since p-value is less than 0.1 significance level, we Reject the null hypothesis.

90% confidence interval for the mean percent infected in the population of all velvetleaf plants is,

(3.45% ,  4.74%)

Since the confidence interval contains all values less than 5%,

The beetle infects less than 5% of seeds, so it is unlikely to be effective in controlling velvet weed.