A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data support the use of a one-sample t test. The relevant hypotheses are H0: µ = 10 versus Ha: µ < 10.(a) If t = -2.4 and = .05 is selected, what conclusion is appropriate?a. Rejectb. Fail to reject(b) If t = -1.83 and = .01 is selected, what conclusion is appropriate?a. Rejectb. Fail to reject(c) If t = 0.57, what conclusion is appropriate?a. Rejectb. Fail to reject

(a) We reject our null hypothesis.

(b) We fail to reject our null hypothesis.

(c) We fail to reject our null hypothesis.

Step-by-step explanation:

We are given that a certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hr.

A random sample of 18 pens is selected.

Let = true average writing lifetime under controlled conditions

So, Null Hypothesis, : 10 hr   {means that the true average writing lifetime under controlled conditions is at least 10 hr}

Alternate Hypothesis, : < 10 hr    {means that the true average writing lifetime under controlled conditions is less than 10 hr}

The test statistics that is used here is one-sample t test statistics;

                           T.S. =  ~

where, = sample mean

             s = sample standard deviation

             n = sample size of pens = 18

          n - 1 = degree of freedom = 18 -1 = 17

Now, the decision rule based on the critical value of t is given by;

(a) Here, test statistics, t = -2.4 and level of significance is 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is less than the critical value of t as -2.4 < -1.74, so we reject our null hypothesis.

(b) Here, test statistics, t = -1.83 and level of significance is 0.01.

Now, at 0.051 significance level, the t table gives critical value of -2.567 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as -2.567 < -1.83, so we fail to reject our null hypothesis.

(c) Here, test statistics, t = 0.57 and level of significance is not given so we assume it to be 0.05.

Now, at 0.05 significance level, the t table gives critical value of -1.74 at 17 degree of freedom.

Here, clearly the value of test statistics is more than the critical value of t as  -1.74 < 0.57, so we fail to reject our null hypothesis.

$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