A random sample of purchases showed the amounts below What

A random sample of purchases showed the amounts ($) below: What is the standard error of the mean? The owner wants to know if the mean purchase amount of all transactions is at least $50. ( What is the null hypothesis? What is the alternative hypothesis and is it one - or two-sided? What is the value of the test statistic ? What is the P-value of the test statistic? Construct a 95 degree % confidence intend for the mean purchases of all customers. Using the confidence interval, what do you conclude at alpha = .05?

Solution

Set Up Hypothesis
Null, H0: U>=50
Alternate, H1: U<50
Test Statistic
Population Mean(U)=50
Sample X(Mean)=49.4788
Standard Deviation(S.D)=21.3404
Number (n)=16
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =49.4788-50/(21.3404/Sqrt(16))
to =-0.098
| to | =0.098
Critical Value
The Value of |t | with n-1 = 15 d.f is 1.753
We got |to| =0.098 & | t | =1.753
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Left Tail -Ha : ( P < -0.0977 ) = 0.46173
Hence Value of P0.05 < 0.46173,Here We Do not Reject Ho

[ANSWERS]
A. Standard Error = (21.3404/Sqrt(16)) = 5.3351
B.
1. NULL: H0: U>=50, H1: U<50
2. One Tailed
3. to =-0.098
4. ( P < -0.0977 ) = 0.46173
5.
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)=49.4788
Standard deviation( sd )=21.3404
Sample Size(n)=16
Confidence Interval = [ 49.4788 ± t a/2 ( 21.3404/ Sqrt ( 16) ) ]
= [ 49.4788 - 2.131 * (5.335) , 49.4788 + 2.131 * (5.335) ]
= [ 38.11,60.848 ]