A very well-known local fast food burger restaurant with several stores is conce

Question

A very well-known local fast food burger restaurant with several stores is concerned with consistency in product and service. Accordingly, management is considering investing in an automated French-fry dispenser to replace the manual dispensing method because the automatic dispenser promises to have more consistent fry cuts versus the manual machine. The machine is expensive and before management determines whether to purchase the new dispensers for all the local stores, a pilot study is conducted in one of the stores with a lower customer count. Although consistency is important, management also wants to make sure the average number of fries produced remains the same as the current manual dispensing method. The average number of fries produced per minute historically was 350 fries. Management samples 36 productions of fries per minute in the pilot study and finds the average number of fries produced is 338 with a standard deviation of 20 fries. You are asked to evaluate the sample results statistically. Complete the following............ A) Perform the appropriate test to determine if the average number of fries produced per minute changed when using the automated machine. Make sure you show all required steps of the hypothesis testing procedure. Use alpha = .05. Show work where necessary - don't just list a single value for your answer.......... B) Construct a 95% confidence interval estimate for the mean number of fries produced with the automated dispenser.......... C) Explain briefly how the confidence interval supports the result of the hypothesis test conducted in part A. D) Provide a one-sentence comment on whether the manager should purchase the new dispenser.

Explanation / Answer

PART A.
Given that,
population mean(u)=350
sample mean, x =338
standard deviation, s =20
number (n)=36
null, Ho: =350
alternate, H1: !=350
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =338-350/(20/sqrt(36))
to =-3.6
| to | =3.6
critical value
the value of |t | with n-1 = 35 d.f is 2.03
we got |to| =3.6 & | t | =2.03
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.6 ) = 0.001
hence value of p0.05 > 0.001,here we reject Ho
ANSWERS
---------------
null, Ho: =350
alternate, H1: !=350
test statistic: -3.6
critical value: -2.03 , 2.03
decision: reject Ho
p-value: 0.001

PART B.
given that,
sample mean, x =338
standard deviation, s =20
sample size, n =36
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 35 d.f is 2.03
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 338 ± Z a/2 ( 20/ Sqrt ( 36) ]
= [ 338-(2.03 * 3.333) , 338+(2.03 * 3.333) ]
= [ 331.233 , 344.767 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 331.233 , 344.767 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean

PART C.
the result from strongly supports that mean value is diffrent from Ho: =350 and confidence we
calculated as similar to above i.e [ 331.233 , 344.767 ] does n't include the value of mean

PART D.
manager should n't purchase the new dispenser as the production of fries is decreased

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