Question

Using Excel.

On its municipal website, the City of Tulsa states that the rate it charges per 5 CCF of residential water is $21.62. How do the residential water rates of other U.S. public utilities compare to Tulsa's rate? The data set contains the rate per 5 CCF of residential water for 42 randomly selected U.S. cities. Formulate hypotheses that can be used to determine whether the population mean rate per 5CCF of residential water charged by U.S. public utilities differs from the $21.62 rate charged by Tulsa. Conduct a T-Test at α = 0.05 to test your hypotheses. Use Excel.

Rate (5 CCF)
10.48
9.18
11.8
6.5
12.42
14.53
15.56
10.12
14.5
16.18
17.6
19.18
17.98
12.85
16.8
17.35
15.64
14.8
18.91
17.99
14.9
18.42
16.05
26.85
22.32
22.76
20.98
23.45
19.05
23.7
19.26
23.75
27.8
27.05
27.14
26.99
24.68
37.86
26.51
39.01
29.46
41.65

1.

Compute the following.
n
df
mean
std dev
std err
critical value
test value
p-value

2. State the null and alternative hypotheses. What type of test is this, right-tailed, left-tailed or two-tailed?

3. Do you reject or fail to reject the null hypothesis. Explain why. Use the α = 0.01 level of significance.  State your conclusion

Answer 1

Given that,
population mean(u)=21.62
standard deviation, σ =7.705
sample mean, x =20.238
number (n)=42
null, Ho: μ=21.62
alternate, H1: μ!=21.62
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 20.238-21.62/(7.705/sqrt(42)
zo = -1.16
| zo | = 1.16
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.16 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.16 ) = 0.25
hence value of p0.05 < 0.25, here we do not reject Ho
ANSWERS
---------------
1.
population mean(u)=21.62
standard deviation, σ =7.705
sample mean, x =20.238
number (n)=42
2.
null, Ho: μ=21.62
alternate, H1: μ!=21.62
test statistic: -1.16
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.2518
we do not have enough evidence to support the claim that whether the population mean rate per 5CCF of residential
water charged by U.S. public utilities differs from the $21.62 rate charged by Tulsa.
3.
Given that,
population mean(u)=21.62
standard deviation, σ =7.705
sample mean, x =20.238
number (n)=42
null, Ho: μ=21.62
alternate, H1: μ!=21.62
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.576
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 20.238-21.62/(7.705/sqrt(42)
zo = -1.16
| zo | = 1.16
critical value
the value of |z α| at los 1% is 2.576
we got |zo| =1.16 & | z α | = 2.576
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.16 ) = 0.25
hence value of p0.01 < 0.25, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=21.62
alternate, H1: μ!=21.62
test statistic: -1.16
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.2518
we do not have enough evidence to support the claim that whether the population mean rate per 5CCF of residential
water charged by U.S. public utilities differs from the $21.62 rate charged by Tulsa.