1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ganezh [65]
3 years ago
11

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

st of 26.31 cents per mile and the sample standard deviation of 2.80 cents per mile. (Assume that operating costs are approximately normally distributed.) Please keep at least 4 significant digits in your answer.
1.Construct a 95% confidence interval for the mean operating cost.2.Construct a 90% confidence interval for the variance of the operating costs.3.The manager wants to believe that the actual mean operating cost is at most 25 cents per mile. Perform the appropriate test at a 5% level of significance. What is the value of your test statistic?What is the critical value? Accept or Reject null hypothesis?4. Test whether the overall standard deviation of the operating costs is more than 2.30 cents per mile or not at a 5% significance level. What is the value of your test statistic? What is the critical value at a 5% significance level? Accept or Reject null hypothesis?
Mathematics
1 answer:
levacccp [35]3 years ago
7 0

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)  

2. 4.7048 \leq \sigma^2 \leq 16.1961

3. t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

t_{crit}=1.753

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

\chi^2 =24.9958

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

\bar X=26.31 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=16-1=15

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that t_{\alpha/2}=2.13

Now we have everything in order to replace into formula (1):

26.31-2.13\frac{2.8}{\sqrt{16}}=24.819    

26.31+2.13\frac{2.8}{\sqrt{16}}=27.801

So on this case the 95% confidence interval would be given by (24.8190;27.8010)  

Part 2

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=16-1=15

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

\chi^2_{\alpha/2}=24.996

\chi^2_{1- \alpha/2}=7.261

And replacing into the formula for the interval we got:

\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}

4.7048 \leq \sigma^2 \leq 16.1961

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:    

Null hypothesis:\mu \leq 25      

Alternative hypothesis:\mu > 25      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Calculate the statistic      

We can replace in formula (1) the info given like this:      

t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

Critical value  

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got t_{crit}=1.753

Conclusion      

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: \sigma \leq 2.3

H1: \sigma >2.3

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be \chi^2 =24.9958

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

You might be interested in
What is the value of the fourth term in geometric sequence for which a1 = 15 and r = 1/3?
Naya [18.7K]
So you express fraction. 
15 x (1/3)^ 3=15/27=5/9
7 0
3 years ago
Read 2 more answers
What is the value of x + |y| when x = –11 and y = –4?                                                                          
NeX [460]
x=-11\\y=-4\\\\x+|y|=\\\\-11+|-4|=\\\\-11+4=-7\\\\Answer\ B.
7 0
3 years ago
PLEASE HELP!!!!
Katena32 [7]

Answer:

C

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Floyd is an aspiring music artist. He has a record contract that pays him a base rate of \$200$200dollar sign, 200 a month and a
Vikentia [17]
$644-$200=444
444 / 12 = 37
8 0
3 years ago
Read 2 more answers
What did the scout say after fixing the little old lady's bicycle horn
Elodia [21]
Did he say your welcome?
 
8 0
3 years ago
Other questions:
  • In a geometric sequence, A1 = 0.3 and r=3.Find A12, to the nearest integer
    15·1 answer
  • Simplify the expression (2 + 6i)(4 - 2i)
    11·1 answer
  • What is the slope of the line shown below?
    6·1 answer
  • PLEASE HELP ME GUYS!!
    11·1 answer
  • Let the cost of a pen be dollars. Write an expression for the total cost of four pens and seven mechanical pencils in terms of .
    14·1 answer
  • The future population of a small european country of 14 million people
    9·1 answer
  • Elliot read a report from a previous year saying that 6%, percent of adults in his city biked to work. He wants to test whether
    10·1 answer
  • 17 times 14 times 123
    5·2 answers
  • Find each angle measure in the figure<br><br><br><br>The angle measures are ___°, ___° And ___°.​
    10·2 answers
  • The raidius of the circle is r=___The equation of the circle in standard form____
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!