Question

The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm. Find the p-value for a test to see whether the candy bars are smaller than they are supposed to be?

Please explain why this is a t or z test and show the p val step in detail with the Excel command used to obtain the answer. Also, if this is norm.s.dist, explain why you can use a standard normal distribution here?

Answer 1

Given that,
population mean(u)=56
standard deviation, σ =0.77
sample mean, x =55.82
number (n)=49
null, Ho: μ=56
alternate, H1: μ<56
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 55.82-56/(0.77/sqrt(49)
zo = -1.64
| zo | = 1.64
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.64 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -1.64 ) = 0.05
hence value of p0.05 < 0.05, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=56
alternate, H1: μ<56
test statistic: -1.64
critical value: -1.645
decision: do not reject Ho
p-value: 0.05
we do not have enough evidence to support the claim that whether the candy bars are smaller than they are supposed to be.