The Z-value is Z=-2.121<Zcritical and the p-value is 0.0169 when percentage of at least 0.8 of the packages must include 3 ounces.
Given that,
A percentage of at least 0.8 of the packages must include 3 ounces or more of chocolate stars for a candy producer to assert that a bridge mix is predominantly chocolate stars. A random sample of 50 packages is tested by quality control to see if the proportion is less than 0.8 at a significance level of 0.05.
We have to determine p, the z-score, and the p-value, use Sheet 6 of the Excel document.
We know that,
In the picture there is a sheet 6.
N=50
n=34
Proportion pbar = n/N=34/50=0.68
H₀: p≥0.8
H₁: p<0.8
Z=pbar-p/√(p(1-p)/N
Z=0.68-0.8/√0.8(1-0.8)/50
Z=-2.121
α=0.05
Reject H₀ if Z<-Zcritical
Zcritical for one tailed (Left) = -1.6449
Z=-2.121<Zcritical
Reject H₀
P(Z < -2.121) = 0.0169
p-value = 0.0169
Since the p-value is less than significance level of 0.05 the null hypothesis is rejected.
There is sufficient data to draw the conclusion that, at the 0.05 level of significance, the proportion of packages containing 3 ounces or more of chocolate stars is less than 0.8.
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