Answer:
Step-by-step explanation:
The hypothesis is written as follows
For the null hypothesis,
µd ≤ 10
For the alternative hypothesis,
µ > 10
This is a right tailed test
Since no population standard deviation is given, the distribution is a student's t.
Since n = 97
Degrees of freedom, df = n - 1 = 97 - 1 = 96
t = (x - µ)/(s/√n)
Where
x = sample mean = 8.9
µ = population mean = 10
s = samples standard deviation = 3.6
t = (8.9 - 10)/(3.6/√97) = - 3
We would determine the p value using the t test calculator. It becomes
p = 0.00172
Since alpha, 0.01 > than the p value, 0.00172, then we would reject the null hypothesis. Therefore, At a 1% level of significance, there is enough evidence that the data do not support the vendor’s claim.
The answer is going to be the second one
Answer: 1.
Yes , the data franchise owner is collecting, will be helpful in order to counter the criticism from the critic. As he has a record for every delivery they are making on daily basis.
2.
The scenario is that for every fifth customer , the owner cross the four lines he makes. Thus a pack of five and easy to remember. As the data on regular week days is less, As can do the same for the third customer. Crossing the 2 lines for 3rd customer.
3.
With this provision of crossing the third customer, the owner can have a better check on the late delivery. As it will bring in his notice earlier as compared to fifth crossing.
4.
With the data as provided in the table, we can come to the conclusion that on weekends , i.e. both on time delivery and late deliveries are high in numbers. It must be because of high demand on that days. So owner must make some provision like hiring some more delivery boys for weekend in order to reduce the late deliveries.
Step-by-step explanation:
Answer:
me :/
Step-by-step explanation:
