I need this questions Answer on Microsoft Excel THank You A

I need this question\'s Answer on Microsoft Excel. THank You.

A courier service advertises that its average delivery time is less than six hours for local deliveries. A random sample of the amount of time this courier takes to deliver packages to an address across town produced the following times (rounded to the nearest hour): 7,3,4, 6,10,5, 6, 4, 3, and 8. Is this sufficient evidence to support the courier\'s advertisement, at the 5 percent level of significance? Find tire 99 percent confidence interval estimate of mean delivery time. What assumption must be made in order to answer these questions?

Solution

1.
Set Up Hypothesis
Null, H0: U=6
Alternate, Average delivery time is less than 6 Hours, H1: U<6
Test Statistic
Population Mean(U)=6
Sample X(Mean)=5.6
Standard Deviation(S.D)=2.271
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =5.6-6/(2.271/Sqrt(9))
to =-0.557
| to | =0.557
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |to| =0.557 & | t | =1.833
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Left Tail -Ha : ( P < -0.557 ) = 0.29556
Hence Value of P0.05 < 0.29556,Here We Do not Reject Ho

We don\'t have the evidence to indicate that Average delivery time is less than 6 Hours

2.
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)=5.6
Standard deviation( sd )=2.271
Sample Size(n)=10
Confidence Interval = [ 5.6 ± t a/2 ( 2.271/ Sqrt ( 10) ) ]
= [ 5.6 - 3.25 * (0.718) , 5.6 + 3.25 * (0.718) ]
= [ 3.266,7.934 ]

3.
Sample is under normal