Question

The Crown Bottling Company has just installed a new bottling process that will fill 16-ounce bottles of the popular Crown Classic Cola soft drink. Both overfilling and underfilling bottles are undesirable: Underfilling leads to customer complaints and overfilling costs the company considerable money. In order to verify that the filler is set up correctly, the company wishes to see whether the mean bottle fill, μ, is close to the target fill of 16 ounces. To this end, a random sample of 37 filled bottles is selected from the output of a test filler run. If the sample results cast a substantial amount of doubt on the hypothesis that the mean bottle fill is the desired 16 ounces, then the filler’s initial setup will be readjusted.

(a) The bottling company wants to set up a hypothesis test so that the filler will be readjusted if the null hypothesis is rejected. Set up the null and alternative hypotheses for this hypothesis test.

H₀ : μ (Click to select)≠= 16 versus H_a : μ (Click to select)≠= 16

(b) Suppose that Crown Bottling Company decides to use a level of significance of α = 0.01, and suppose a random sample of 37 bottle fills is obtained from a test run of the filler. For each of the following four sample means— x¯x¯ = 16.06, x¯x¯ = 15.96, x¯x¯ = 16.03, and x¯x¯ = 15.93 — determine whether the filler’s initial setup should be readjusted. In each case, use a critical value, a p-value, and a confidence interval. Assume that σ equals .1. (Round your z to 2 decimal places and p-value to 4 decimal places and CI to 3 decimal places.)

x¯x¯ = 16.06

x¯x¯⁢ = 15.96

x¯x¯⁢ = 16.03

x¯x¯⁢ = 15.93

Answer 1

Given that,
population mean(u)=16
standard deviation, sigma =0.1
number (n)=37
null, Ho: μ=16
alternate, H1: μ!=16
level of significance, alpha = 0.01
from standard normal table, two tailed z alpha/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))
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a. assume sample mean, x =16.06
zo = 16.06-16/(0.1/sqrt(37)
zo = 3.6497
| zo | = 3.6497
critical value
the value of |z alpha| at los 1% is 2.576
we got |zo| =3.6497 & | z alpha | = 2.576
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 3.6497 ) = 0.0003
hence value of p0.01 > 0.0003, here we reject Ho
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b. assume sample mean, x =15.96
zo = 15.96-16/(0.1/sqrt(37)
zo = -2.4331
| zo | = 2.4331
critical value
the value of |z alpha| at los 1% is 2.576
we got |zo| =2.4331 & | z alpha | = 2.576
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.4331 ) = 0.015
hence value of p0.01 < 0.015, here we do not reject Ho
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c. assume sample mean, x =16.03
zo = 16.03-16/(0.1/sqrt(37)
zo = 1.8248
| zo | = 1.8248
critical value
the value of |z alpha| at los 1% is 2.576
we got |zo| =1.8248 & | z alpha | = 2.576
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.8248 ) = 0.068
hence value of p0.01 < 0.068, here we do not reject Ho
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d. assume sample mean, x =15.93
zo = 15.93-16/(0.1/sqrt(37)
zo = -4.2579
| zo | = 4.2579
critical value
the value of |z alpha| at los 1% is 2.576
we got |zo| =4.2579 & | z alpha | = 2.576
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -4.2579 ) = 0
hence value of p0.01 > 0, here we reject Ho