Question

the manager of a candy supply store, and you want to determine whether the mean amount of candy contained in 5-ounce bag purchased from a nationally known refiner is actually 5 ounces. You know from the refiner's specifications that the amount of candy per bag is normally distributed with the standard deviation of 0.25 ounces. You select a random sample of 30 bags, and the mean amount of candy per 5-ounce bag is 5.10 ounces.

What are the hypotheses that you would use to conduct a hypothesis test to see if the population mean amount is different from 5 ounces?

h1:

b. State the appropriate test used and reason.

c. Find the critical value(s) of the test statistic at the 0.01 level of significance. Keep at least 2 decimal places. Show work. A loss of marks will result for not showing work even if your answer is correct.

Critical Value(s)

d. Construct a 99% confidence interval estimate for the population mean amount of candy. Fill in the table below. Interpret your result keeping in mind that you do not know population mean. Keep at least 4 decimal places. Show work. A loss of marks will result for not showing work even if your answer is correct

Confidence Interval

Interval Lower Limit

Interval Upper Limit

e. Based on your answer in part d, what conclusions do you reach? Explain by providing the decision with reason.

f. Suppose the population mean is indeed 5 ounce. What is the probability that the sample mean amount of candy is between 5.08 and 4.95 ounces from a random sample of 30 bags? Keep at least 4 decimal places. Show work. A loss of marks will result for not showing work even if your answer if correct

Probability is:

Answer 1

a.

Given that,
population mean(u)=5
standard deviation, σ =0.25
sample mean, x =5.1
number (n)=30
null, Ho: μ=5
alternate, H1: μ!=5
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 5.1-5/(0.25/sqrt(30)
zo = 2.191
| zo | = 2.191
critical value
the value of |z α| at los 1% is 2.576
we got |zo| =2.191 & | z α | = 2.576
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.191 ) = 0.028
hence value of p0.01 < 0.028, here we do not reject Ho
ANSWERS
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b.
Z test for single mean with known standard deviation
null, Ho: μ=5
alternate, H1: μ!=5
c.
test statistic: 2.191
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.028
we do not have enough evidence to support the claim that if the population mean amount is different from 5 ounces
d.
TRADITIONAL METHOD
given that,
standard deviation, σ =0.25
sample mean, x =5.1
population size (n)=30
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 0.25/ sqrt ( 30) )
= 0.0456
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 0.0456
= 0.1176
III.
CI = x ± margin of error
confidence interval = [ 5.1 ± 0.1176 ]
= [ 4.9824,5.2176 ]
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DIRECT METHOD
given that,
standard deviation, σ =0.25
sample mean, x =5.1
population size (n)=30
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 5.1 ± Z a/2 ( 0.25/ Sqrt ( 30) ) ]
= [ 5.1 - 2.576 * (0.0456) , 5.1 + 2.576 * (0.0456) ]
= [ 4.9824,5.2176 ]
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e.
interpretations:
1. we are 99% sure that the interval [4.9824 , 5.2176 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
f.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 5
standard Deviation ( sd )= 0.25/ Sqrt ( 30 ) =0.0456
sample size (n) = 30
he probability that the sample mean amount of candy is between 5.08 and 4.95 ounces from a random sample of 30 bags
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 4.95) = (4.95-5)/0.25/ Sqrt ( 30 )
= -0.05/0.04564
= -1.09545
= P ( Z <-1.09545) From Standard Normal Table
= 0.13666
P(X < 5.08) = (5.08-5)/0.25/ Sqrt ( 30 )
= 0.08/0.04564 = 1.75271
= P ( Z <1.75271) From Standard Normal Table
= 0.96017
P(4.95 < X < 5.08) = 0.96017-0.13666 = 0.8235