Question

B??

Data on the weights (b) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal Complete parts (a) and (b) below. Use a 0.05 significance level for both parts. Diet 38 0.78178 lb 0.00443 lb μ2 38 0.80874 Ib 0.00741 lb a. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda. What are the null and alternative hypotheses? H1 : μ1 > P2 The test statistic, t, is -19.25. (Round to two decimal places as needed.) The P-value is 0.000. (Round to three decimal places as needed.) State the conclusion for the test. O A 。B. Re ect the null hypothesis. There is not sufficient evidence to support the daim that the cans of diet soda have mean we hts that are lower than the mean weight fo the regular soda. Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the cans o diet soda hawe mean weights that are lower than the mean weight or the e ular soda. C. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. D. Reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda. b. Construct a confidence interval appropriate for the hypothesis test in part (a). (Round to three decimal places as needed.)

Answer 1

a.
Given that,
mean(x)=0.78178
standard deviation , s.d1=0.00443
number(n1)=38
y(mean)=0.80874
standard deviation, s.d2 =0.00741
number(n2)=38
null, Ho: u1 =u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.687
since our test is left-tailed
reject Ho, if to < -1.687
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.78178-0.80874/sqrt((0.00002/38)+(0.00005/38))
to =-19.25
| to | =19.25
critical value
the value of |t α| with min (n1-1, n2-1) i.e 37 d.f is 1.687
we got |to| = 19.25029 & | t α | = 1.687
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -19.2503 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -19.25
critical value: -1.687
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that cans of diet soda have mean weights lower than mean weights of regular soda.

b.
TRADITIONAL METHOD
given that,
mean(x)=0.78178
standard deviation , s.d1=0.00443
number(n1)=38
y(mean)=0.80874
standard deviation, s.d2 =0.00741
number(n2)=38
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((0/38)+(0/38))
= 0.001
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 37 d.f is 2.026
margin of error = 2.026 * 0.001
= 0.003
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (0.78178-0.80874) ± 0.003 ]
= [-0.03 , -0.024]
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DIRECT METHOD
given that,
mean(x)=0.78178
standard deviation , s.d1=0.00443
sample size, n1=38
y(mean)=0.80874
standard deviation, s.d2 =0.00741
sample size,n2 =38
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 = [( 0.78178-0.80874) ± t a/2 * sqrt((0/38)+(0/38)]
= [ (-0.027) ± t a/2 * 0.001]
= [-0.03 , -0.024]
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interpretations:
1. we are 95% sure that the interval [-0.03 , -0.024] 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