The telephone company was interested in measuring the averag

The telephone company was interested in measuring the average daily usage in minutes for household telephones in a specific area in order to determine whether this rate is different from a state-wide average daily usage for households. Suppose that a random sample of nine households were sampled on random days, producing the following times (in minutes): 35, 59, 42, 44, 31, 46, 24, 56, 50 a. Estimate the average daily usage using a 90% confidential interval. ( Term 1 , Term 2 ) b. Suppose that the state-wide average for households has been found to be 45 minutes. Does this data provide evidence to indicate that the average in this area differs from the state-wide average? Use x =0.05

Step 1

Ho:M =

H1:M =/ (does not equal)

Step 2

df =

Step 3 t =

Step 4 critical t =

Step 5 Can we reject Ho or do we fail to reject Ho?

Solution

a)
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)=43
Standard deviation( sd )=11.478
Sample Size(n)=9
Confidence Interval = [ 43 ± t a/2 ( 11.478/ Sqrt ( 9) ) ]
= [ 43 - 1.86 * (3.826) , 43 + 1.86 * (3.826) ]
= [ 35.884,50.116 ]

b)
Set Up Hypothesis
Null, No significance b/w them H0: U=45
Alternate, H1: U!=45
Test Statistic
Population Mean(U)=45
Sample X(Mean)=43
Standard Deviation(S.D)=11.478
Number (n)=9
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =43-45/(11.478/Sqrt(8))
to =-0.523
| to | =0.523
Critical Value
The Value of |t | with n-1 = 8 d.f is 2.306
We got |to| =0.523 & | t | =2.306
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != -0.5227 ) = 0.6153
Hence Value of P0.05 < 0.6153,Here We Do not Reject Ho