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 39 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.

b)

Suppose that Crown Bottling Company decides to use a level of significance of α = 0.01, and suppose a random sample of 39 bottle fills is obtained from a test run of the filler. For each of the following four sample means— x¯x¯ = 16.05, x¯x¯ = 15.95, x¯x¯ = 16.03, and x¯x¯ = 15.97 — 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.)

what is the

i) Z ii)

P-Value and

iii) Confidence Internal

x¯x¯ = 16.05

Answer 1

a) Null and Alternative hypothesis:  

Ho : µ =    16

H1 : µ ≠    16

b)

x̅ = 16.05, σ = 0.1, n = 39  

Test statistic:  

z = (x̅- µ)/(σ/√n) = (16.05 - 16)/(0.1/√39) =    3.12

p-value = 2*(1-NORM.S.DIST(ABS(3.1225, 1) =    0.0018

Decision:  

p-value < α, Reject the null hypothesis     
  
99% Confidence interval :  

At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) =    2.576

Lower Bound = x̅ - z_c*σ/√n = 16.05 - 2.576 * 0.1/√39 =    16.009

Upper Bound = x̅ + z_c*σ/√n = 16.05 + 2.576 * 0.1/√39 =    16.091

----

x̅ = 15.95, σ = 0.1, n = 39

Test statistic:  

z = (x̅- µ)/(σ/√n) = (15.95 - 16)/(0.1/√39) =    -3.12

p-value :  

p-value = 2*(1-NORM.S.DIST(ABS(-3.1225, 1) =    0.0018

Decision:  

p-value < α, Reject the null hypothesis  

99% Confidence interval :  

At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) =    2.576

Lower Bound = x̅ - z_c*σ/√n = 15.95 - 2.576 * 0.1/√39 =    15.909

Upper Bound = x̅ + z_c*σ/√n = 15.95 + 2.576 * 0.1/√39 = 15.991

-----

x̅ = 16.03, σ = 0.1, n = 39  

Test statistic:  

z = (x̅- µ)/(σ/√n) = (16.03 - 16)/(0.1/√39) =    1.87

p-value :  

p-value = 2*(1-NORM.S.DIST(ABS(1.8735, 1) =    0.0610

Decision:  

p-value > α, Do not reject the null hypothesis  

99% Confidence interval :  

At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) =    2.576

Lower Bound = x̅ - z_c*σ/√n = 16.03 - 2.576 * 0.1/√39 =    15.989

Upper Bound = x̅ + z_c*σ/√n = 16.03 + 2.576 * 0.1/√39 =    16.071

------

x̅ = 15.97, σ = 0.1, n = 39  

Test statistic:  

z = (x̅- µ)/(σ/√n) = (15.97 - 16)/(0.1/√39) = -1.87

p-value :  

p-value = 2*(1-NORM.S.DIST(ABS(-1.8735, 1) =    0.0610

Decision:  

p-value > α, Do not reject the null hypothesis  

99% Confidence interval :  

At α = 0.01 two tailed critical value, z_c = ABS(NORM.S.INV(0.01/2)) =    2.576

Lower Bound = x̅ - z_c*σ/√n = 15.97 - 2.576 * 0.1/√39 =    15.929

Upper Bound = x̅ + z_c*σ/√n = 15.97 + 2.576 * 0.1/√39 =    16.011