PuYouChen Company is promoting a food to a ‘JingGang Supermarket’. They want to

Question

PuYouChen Company is promoting a food to a ‘JingGang Supermarket’. They want to convince the supermarket manager that the food is worth stocking. The manager has agreed to stock if it sells well enough, at least =50 units/week. PuYouChen arranges the JingGang Supermarket to conduct a 36 week trial with the goal of convincing the supermarket to stock the food. They find that the supermarket has sold an average of 55 units per week over that period with a standard deviation of s = 6.99.

(a). Compute a 95% confidence interval for , the mean number of food the supermarket will sell per week in the long run. Does it look like the supermarket will be willing to stock the food? Explain.

(b). Perform the hypothesis test using test statistics that PuYouChen will conduct to convince the supermarket the food is worth stocking. In other words, give the null and alternative hypotheses, both mathematically and in words, and explain your reasoning.

(c). Conduct the hypothesis test using p-value and state your real-world conclusions assuming you need to be 95% sure the food is worthwhile before you stock it.. Is this consistent with your result from part (b)?

(d). Suppose that PuYouChen makes a $5 profit on every unit sold by the supermarket. Furthermore, suppose they need to average profits of at least $260 per week for it to be worthwhile to market the food at that supermarket. Can they be 95% (really 97.5%) sure of meeting their goal? Explain. (Hint: Use the CI computed in part (a)).

Explanation / Answer

a.
Confidence Interval
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
Mean(x)=55
Standard deviation( sd )=6.99
Sample Size(n)=36
Confidence Interval = [ 55 ± t a/2 ( 6.99/ Sqrt ( 36) ) ]
= [ 55 - 2.03 * (1.165) , 55 + 2.03 * (1.165) ]
= [ 52.635,57.365 ]
b.
Given that,
population mean(u)=50
sample mean, x =55
standard deviation, s =6.99
number (n)=36
null, Ho: =50
alternate, H1: !=50
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 =55-50/(6.99/sqrt(36))
to =4.292
| to | =4.292
critical value
the value of |t | with n-1 = 35 d.f is 2.03
we got |to| =4.292 & | 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 != 4.2918 ) = 0.0001
hence value of p0.05 > 0.0001,here we reject Ho
ANSWERS
---------------
null, Ho: =50
alternate, H1: !=50
test statistic: 4.292
critical value: -2.03 , 2.03
decision: reject Ho
c.

p-value: 0.0001

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