A city councilwoman is concerned that the new contractor she hired is taking too long to replace defective streetlights. She would like to perform a hypothesis test to determine if the replacement time for streetlights under the new contractor is in fact longer than the replacement time under the previous contractor, which was 3.2 days on average. She selects a random sample of 12 streetlight service calls and obtains the following times to replacement (in days). Use a significance level of 0.05 6.2 7.15.4 5.57.52.6 4.3 2.9 3.7 0.7 5.6 1.7

Required:
a. Define μ in the context of the problem and state the appropriate hypotheses.
b. State and check the conditions for a hypothesis test of the mean. Since the sample size is smaller than 30, you will need to calculate a normal probability plot to determine if the data come from a normal distribution.

There is significant evidence to conclude that the replacement time for streetlights under the new contractor is longer than the replacement time under the previous contractor.

Step-by-step explanation:

Given the data :

6.2 7.1 5.4 5.5 7.5 2.6 4.3 2.9 3.7 0.7 5.6 1.7

The hypothesis:

H0: μ = 3.2

H1 : μ > 3.2

n = sample size = 12

The sample mean, xbar = ΣX / n

xbar = 53.2 / 12

xbar = 4.43

Using calculator;

Sample standard deviation, s = 2.147

The test statistic :

(xbar - μ) ÷ (s/sqrt(n))

(4.43 - 3.2) ÷ (2.147/sqrt(12)

1.23 / 0.6197855

Test statistic = 1.985

The Pvalue using the Pvalue from Tscore calculator :

Tscore = 1.985 ; df = 12 - 1 = 11

Pvalue = 0.036

Since Pvalue < α ; We reject the Null

Hence, we conclude that the replacement time for streetlights under the new contractor is longer than the replacement time under the previous contractor.

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40