Question

Do you get more for your money by buying loose or prepackaged carrots? 40 random packages of carrots, labeled “1 pound” were selected from 8 different grocery stores and weighed to the nearest gram. (Note: grams were used as the unit of measurement so that the measurement could be more precise. One pound is equivalent to 454 grams.) The sample mean weight was 512.2 grams with a standard deviation of 39.3 grams. Is there evidence that the mean weight of a “1 pound” bag of carrots weighs differently than 454 grams (the equivalent of 1 pound)

a. Define the parameter(s) and state the hypotheses.

b. What would be a good graph to display the data from this study?

c. Are the conditions for the appropriate analysis met? Explain and show work or attach a graph if appropriate.

d. Compute the appropriate test statistic (show work). (3 points)

e. How many degrees of freedom does the test statistic have?

f. Find the p-value.

g. Based on your p-value, what do you conclude? Use  = 0.01.

h. Construct a 90% confidence interval for the mean weight (in grams) of a “1 pound” bag of carrots. Write a sentence interpreting your confidence interval.

Answer 1

Given that,
population mean(u)=454
sample mean, x =512.2
standard deviation, s =39.3
number (n)=40
null, Ho: μ=454
alternate, H1: μ!=454
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.708
since our test is two-tailed
reject Ho, if to < -2.708 OR if to > 2.708
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =512.2-454/(39.3/sqrt(40))
to =9.3661
| to | =9.3661
critical value
the value of |t α| with n-1 = 39 d.f is 2.708
we got |to| =9.3661 & | t α | =2.708
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 9.3661 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
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a.
null, Ho: μ=454
alternate, H1: μ!=454
c.
yes,
the conditions for the appropriate analysis met
d.
test statistic: 9.3661
e.
d.f is 2.708
f.
critical value: -2.708 , 2.708
decision: reject Ho
p-value: 0
g.
we have enough evidence to support the claim that the mean weight of a “1 pound” bag of carrots weighs differently than 454 grams.
h.
TRADITIONAL METHOD
given that,
sample mean, x =512.2
standard deviation, s =39.3
sample size, n =40
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 39.3/ sqrt ( 40) )
= 6.214
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 1.685
margin of error = 1.685 * 6.214
= 10.47
III.
CI = x ± margin of error
confidence interval = [ 512.2 ± 10.47 ]
= [ 501.73 , 522.67 ]
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DIRECT METHOD
given that,
sample mean, x =512.2
standard deviation, s =39.3
sample size, n =40
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 1.685
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 512.2 ± t a/2 ( 39.3/ Sqrt ( 40) ]
= [ 512.2-(1.685 * 6.214) , 512.2+(1.685 * 6.214) ]
= [ 501.73 , 522.67 ]
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interpretations:
1) we are 90% sure that the interval [ 501.73 , 522.67 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean