Question

A retailer discovers that 3 jars from his last shipment of Spiffy peanut butter contained between 15.85 and 15.92 oz of peanut butter, despite the labeling indicating that each jar should contain 16 oz. of peanut butter. He is wondering if Spiffy is cheating its customers by filling its jars with less product than advertised. He decides to measure the weight of 50 jars from the shipment and use hypothesis testing to verify this.

1. (f) In his sample of 50 jars, the retailer finds an average weight of 15.84 oz and a sample standard deviation of 0.5 oz. He decides to use a significance level of 0.04. What is the conclusion from this hypothesis testing? Can you conclude that Spiffy is cheating its customers?

2. (g) What is the p-value? What is the meaning of this number?

3. (h) For what values of the sample mean would the null hypothesis be rejected?

4. (i) Calculate the probability of type II error if the true mean is 15.7 oz.

5. (j) Solve (f), (h) and (i) when the level of significance is 0.01. Is your new answer for (f) consistent with the p-value found in (g)? How is the probability of type II error affected when the probability of type I error changes?

Answer 1

Given that,
population mean(u)=15.84
standard deviation, σ =0.5
sample mean, x =15.7
number (n)=50
null, Ho: μ=15.84
alternate, H1: μ!=15.84
level of significance, α = 0.04
from standard normal table, two tailed z α/2 =2.054
since our test is two-tailed
reject Ho, if zo < -2.054 OR if zo > 2.054
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 15.7-15.84/(0.5/sqrt(50)
zo = -1.98
| zo | = 1.98
critical value
the value of |z α| at los 4% is 2.054
we got |zo| =1.98 & | z α | = 2.054
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 != -1.98 ) = 0.048
hence value of p0.04 < 0.048, here we do not reject Ho
ANSWERS
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f.
null, Ho: μ=15.84
alternate, H1: μ!=15.84
test statistic: -1.98
critical value: -2.054 , 2.054
decision: do not reject Ho
g.
p-value: 0.048
h.
we do not have enough evidence to support the claim that average weight of 15.84 oz.
i.
Given that,
Standard deviation, σ =0.5
Sample Mean, X =15.7
Null, H0: μ=15.84
Alternate, H1: μ!=15.84
Level of significance, α = 0.04
From Standard normal table, Z α/2 =2.0537
Since our test is two-tailed
Reject Ho, if Zo < -2.0537 OR if Zo > 2.0537
Reject Ho if (x-15.84)/0.5/√(n) < -2.0537 OR if (x-15.84)/0.5/√(n) > 2.0537
Reject Ho if x < 15.84-1.0269/√(n) OR if x > 15.84-1.0269/√(n)
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Suppose the size of the sample is n = 50 then the critical region
becomes,
Reject Ho if x < 15.84-1.0269/√(50) OR if x > 15.84+1.0269/√(50)
Reject Ho if x < 15.6948 OR if x > 15.9852
Implies, don't reject Ho if 15.6948≤ x ≤ 15.9852
Suppose the true mean is 15.7
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(15.6948 ≤ x ≤ 15.9852 | μ1 = 15.7)
= P(15.6948-15.7/0.5/√(50) ≤ x - μ / σ/√n ≤ 15.9852-15.7/0.5/√(50)
= P(-0.0735 ≤ Z ≤4.0333 )
= P( Z ≤4.0333) - P( Z ≤-0.0735)
= 1 - 0.4707 [ Using Z Table ]
= 0.5293
For n =50 the probability of Type II error is 0.5293

j.
Given that,
population mean(u)=15.84
standard deviation, σ =0.5
sample mean, x =15.7
number (n)=50
null, Ho: μ=15.84
alternate, H1: μ!=15.84
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 = 15.7-15.84/(0.5/sqrt(50)
zo = -1.98
| zo | = 1.98
critical value
the value of |z α| at los 1% is 2.576
we got |zo| =1.98 & | 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 != -1.98 ) = 0.048
hence value of p0.01 < 0.048, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=15.84
alternate, H1: μ!=15.84
test statistic: -1.98
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.048
we do not have enough evidence to support the claim that average weight is 15.84
here also
we fails to reject the null hypothesis so that type 2 error is possible.
type 1 error is possible only it reject the null hypothesis.
Given that,
Standard deviation, σ =0.5
Sample Mean, X =15.7
Null, H0: μ=15.84
Alternate, H1: μ!=15.84
Level of significance, α = 0.01
From Standard normal table, Z α/2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-15.84)/0.5/√(n) < -2.5758 OR if (x-15.84)/0.5/√(n) > 2.5758
Reject Ho if x < 15.84-1.2879/√(n) OR if x > 15.84-1.2879/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 50 then the critical region
becomes,
Reject Ho if x < 15.84-1.2879/√(50) OR if x > 15.84+1.2879/√(50)
Reject Ho if x < 15.6579 OR if x > 16.0221
Implies, don't reject Ho if 15.6579≤ x ≤ 16.0221
Suppose the true mean is 15.7
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(15.6579 ≤ x ≤ 16.0221 | μ1 = 15.7)
= P(15.6579-15.7/0.5/√(50) ≤ x - μ / σ/√n ≤ 16.0221-15.7/0.5/√(50)
= P(-0.5954 ≤ Z ≤4.5552 )
= P( Z ≤4.5552) - P( Z ≤-0.5954)
= 1 - 0.2758 [ Using Z Table ]
= 0.7242
For n =50 the probability of Type II error is 0.7242