On average, “jumbo” bags of potato chips produced at a particular potato chip fa

Question

On average, “jumbo” bags of potato chips produced at a particular potato chip factory are supposed to weigh 32 ounces. The morning-shift supervisor at this factory suspects that the night-shift workers underfill the bags (and eat what they are not packing). The morning-shift supervisor plans to investigate by weighing a random sample of bags of potato chips filled on the night shift.

a) What are the null and alternative hypotheses involved in the test? Your hypotheses should involve one µ; state explicitly what that µ represents.

b) The sample size (use sample size for part b), sample mean, and sample standard deviation of the random sample of bags weighed by the supervisor are shown in the table above. (i) Calculate a t statistic, being sure to say how many degrees of freedom the statistic has and (ii) find the P value for the test of the hypotheses you specified in part a, saying whether the test is one-sided or two-sided. (An approximate P value is fine [e.g., .2 < P < .25].) (Make sure you do each of these two things; labeling your answers as i and ii is recommended.)

c) The supervisor feels that she should have good evidence of pilferage on the night shift before making accusations against the night-shift workers, and so sets = .01. (i) Is the result (from part b) statistically significant at = .01? (Answer “yes” or “no”.) (ii) Explain your answer by describing the relationship of P and .01 (i.e., >, <, =, , etc.). (Be sure you answer each of these questions.)

d) Suppose that the morning shift supervisor had conducted this investigation with a larger sample (see the table for the sample size for part d) and, with this larger sample, had found precisely the same sample mean and sample standard deviation. (i) Calculate a t statistic, including a statement of how many degrees of freedom the statistic has; (ii) find the P value, saying whether the test is one-sided or two-sided; and (iii) say whether the result is statistically significant at = .01. (An approximate P value is fine [e.g., .2 < P < .25].) (Make sure you do each of these three things.)

e) If the morning shift supervisor draws different conclusions in parts b/c and d (and she should!), explain very briefly why there is a discrepancy (given that the sample mean is identical in the two tests).

Explanation / Answer

Given that,
population mean(u)=32
standard deviation, =1.5
sample mean, x =30.4
number (n)=5
null, Ho: >=32
alternate, H1: <32
level of significance, = 0.01
from standard normal table,left tailed z /2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 30.4-32/(1.5/sqrt(5)
zo = -2.38514
| zo | = 2.38514
critical value
the value of |z | at los 1% is 2.326
we got |zo| =2.38514 & | z | = 2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -2.38514 ) = 0.00854
hence value of p0.01 > 0.00854, here we reject Ho
ANSWERS
---------------
a.
null, Ho: >=32
alternate, H1: <32
b.
i.
test statistic: -2.38514
ii.
critical value: -2.326
decision: reject Ho
p-value: 0.00854
c.
i.
yes,from part:b result is statistically significant at = .01
ii.
At statistically significant at = .01
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -2.38514 ) = 0.00854
hence value of p0.01 > 0.00854, here we reject Ho
and
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.38514 ) = 0.01707
hence value of p0.01 < 0.01707, here we do not reject Ho
d.
sample size now n=14 increses
Given that,
population mean(u)=32
standard deviation, =1.5
sample mean, x =30.4
number (n)=14
null, Ho: =32
alternate, H1: !=32
level of significance, = 0.01
from standard normal table, two tailed z /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))
zo = 30.4-32/(1.5/sqrt(14)
zo = -3.9911
| zo | = 3.9911
critical value
the value of |z | at los 1% is 2.576
we got |zo| =3.9911 & | z | = 2.576
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -3.9911 ) = 0.00007
hence value of p0.01 > 0.00007, here we reject Ho
ANSWERS
---------------
i.
null, Ho: =32
alternate, H1: !=32
test statistic: -3.9911
ii.
critical value: -2.576 , 2.576
decision: reject Ho
p-value: 0.00007
iii.
yes,from result is statistically significant at = .01
e.
above all parts sample mean is identical and result is statistically significant at = .01.
conclusions are also diffrent because it depend on the sample size and two tailed,left tailed test.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.