Question

A union of restaurant and foodservice workers would like to estimate this year's mean hourly wage u for foodservice workers in the U.S. Last year's mean hourly wage was $8.25, and there is reason to believe that this year's value is greater than last year's The union decides to do a statistical test to see if it can be concluded that the mean has increased. The union chooses a random sample of 125 wages from this year Suppose that the population of hourly wages of foodservice workers in the U.S. has a standard deviation of $1.15 and that the union performs its hypothesis test using the 0.01 level of significance. Based on this information, answer the questions below. Carry your intermediate computations to at least four decimal places, and round your responses as indicated. (If necessary, consult a list of formulas.) His > What are the null and alternative hypotheses that the union should use for the test? 2 2 Assuming that the actual value of us $8.59, what is the probability that the union accepts the null hypothesis? Round your response to at least two decimal places 0 What is the probability that the union rejects the roll lvpothesis when, in fact, it is true?round your response to at least two decimal places The Dow the second store the one

Answer 1

Given that,
population mean(u)=8.25
standard deviation, σ =1.15
sample mean, x =8.59
number (n)=125
null, Ho: μ=8.25
alternate, H1: μ>8.25
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 8.59-8.25/(1.15/sqrt(125)
zo = 3.305
| zo | = 3.305
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =3.305 & | z α | = 2.326
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : right tail - ha : ( p > 3.305 ) = 0
hence value of p0.01 > 0, here we reject Ho
ANSWERS
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null, Ho: μ=8.25
alternate, H1: μ>8.25
test statistic: 3.305
critical value: 2.326
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that this year's value is greater than last year's value.
i.
Given that,
Standard deviation, σ =1.15
Sample Mean, X =8.59
Null, H0: μ=8.25
Alternate, H1: μ>8.25
Level of significance, α = 0.01
From Standard normal table, Z α/2 =2.3263
Since our test is right-tailed
Reject Ho, if Zo < -2.3263 OR if Zo > 2.3263
Reject Ho if (x-8.25)/1.15/√(n) < -2.3263 OR if (x-8.25)/1.15/√(n) > 2.3263
Reject Ho if x < 8.25-2.6752/√(n) OR if x > 8.25-2.6752/√(n)
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Suppose the size of the sample is n = 125 then the critical region
becomes,
Reject Ho if x < 8.25-2.6752/√(125) OR if x > 8.25+2.6752/√(125)
Reject Ho if x < 8.0107 OR if x > 8.4893
Implies, don't reject Ho if 8.0107≤ x ≤ 8.4893
Suppose the true mean is 8.59
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(8.0107 ≤ x ≤ 8.4893 | μ1 = 8.59)
= P(8.0107-8.59/1.15/√(125) ≤ x - μ / σ/√n ≤ 8.4893-8.59/1.15/√(125)
= P(-5.632 ≤ Z ≤-0.979 )
= P( Z ≤-0.979) - P( Z ≤-5.632)
= 0.1638 - 0 [ Using Z Table ]
= 0.1638
For n =125 the probability of Type II error is 0.1638
ii.
level of significance =0.10
Given that,
Standard deviation, σ =1.15
Sample Mean, X =8.59
Null, H0: μ=8.25
Alternate, H1: μ>8.25
Level of significance, α = 0.1
From Standard normal table, Z α/2 =1.2816
Since our test is right-tailed
Reject Ho, if Zo < -1.2816 OR if Zo > 1.2816
Reject Ho if (x-8.25)/1.15/√(n) < -1.2816 OR if (x-8.25)/1.15/√(n) > 1.2816
Reject Ho if x < 8.25-1.4738/√(n) OR if x > 8.25-1.4738/√(n)
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Suppose the size of the sample is n = 125 then the critical region
becomes,
Reject Ho if x < 8.25-1.4738/√(125) OR if x > 8.25+1.4738/√(125)
Reject Ho if x < 8.1182 OR if x > 8.3818
Implies, don't reject Ho if 8.1182≤ x ≤ 8.3818
Suppose the true mean is 8.59
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(8.1182 ≤ x ≤ 8.3818 | μ1 = 8.59)
= P(8.1182-8.59/1.15/√(125) ≤ x - μ / σ/√n ≤ 8.3818-8.59/1.15/√(125)
= P(-4.5869 ≤ Z ≤-2.0241 )
= P( Z ≤-2.0241) - P( Z ≤-4.5869)
= 0.0215 - 0 [ Using Z Table ]
= 0.0215
For n =125 the probability of Type II error is 0.0215
power of the second test is lower than power of the first test.