Question

PRINTER VER B ACK NEXT ASSIGNMENT RESOURCES One Rooney Chapter 4. Problem 06 Chapter 4. Probleme RS Review Results by Shet Chapter 4, Problem 068 The sugar content of the syrup in canned peaches is normally distributed, and the variance is thought to be o2 = 18 (mg) (a) Test the hypothesis that the variance is not 18 (mg)2 if a random sample of n = 10 cans yields a sample standard deviation of s = 4.5 mg, using a fixed-level test with a = 0.05. State any necessary assumptions about the underlying distribution of the data. Round your answers to 3 decimal places. (b) What is the P-value for this test? Round your answers to 2 decimal places (c) Find a 95% two-sided Cl for o. Round your answers to 3 decimal places (d) Use the Cl in part (c) to test the hypothesis Ho b e rejected 6) Pake- The nul typothes be rejected LA P P 00.00 LVL All Rights Reserved. A Dision of the

Please show the steps for all parts

Answer 1

Here we have given that

n=sample size=10

s= sample standard deviation =3.5 mg

The chi-square test for variance require that the orignal population be normally distributed and here, we we have given that the sugar content of the syrup in canned peaches is normally distributed so, here normality of population assumption is satisfied.

Claim: To check weather the popuation variance is not 18
(mg)
.

The null and alternative hypothesis are

Ho: 02 = 18
  
(mg)

v/s

$H1: \sigma ^2 \ne 18$
(mg)

Now, we can find teh test statistic,

$\chi^2$-Statistics =
(n-1) S2

   =$\frac{(10-1)4.5^2}{18}$

=10.13

we get the test statistics is 10.13

(B)

Now we find the p-value

$\alpha$= level of significance=0.05

Degrees of freedom = n-1 = 10-1 =9

P-value=0.3401 using EXCEL = CHIDIST(  $\chi^2$-Statistics=10.13,D.F=9)

Decision:

Here, p-value > 0.05 $\alpha$

We fail to reject the Null hypothesis Ho

Conclusion:

We can conclude that there is not sufficient evidence to suppor the claim the popuation variance is not 18
(mg)
.

i.e. here population variance is 18
(mg)
.

(C)

Now we want to find the 95% confidence interval for population standard deviation $\sigma$

$\sqrt {\frac{(n-1)S^2}{\chi ^2 \frac{\alpha}{2}}} < \sigma < \sqrt {\frac{(n-1)S^2}{\chi ^2 1- \frac{\alpha}{2}}}$

1st we calcualte the critical values

c= confidence level = 0.95

$\alpha$= level of significance = 1-c = 1-0.95 = 0.05

Degress of freedom = n-1 = 10-1=9

Using EXCEL we can find this command is

=CHIINV(Probablity, D.F)

$\chi ^2_{R}$=$\chi ^2(\frac{\alpha}{2},D.F)$= $\chi ^2(\frac{0.05}{2},9)$ =$\chi ^2(0.025,9)$=19.02

$\chi ^2_{L}$=$\chi ^2 (1-\frac{\alpha}{2}, D.F)$=$\chi ^2 (1-\frac{0.05}{2}, 9)$=$\chi ^2 (0.975, 9)$ =2.70

Now,

$\sqrt {\frac{(n-1)S^2}{\chi ^2 \frac{\alpha}{2}}} < \sigma < \sqrt {\frac{(n-1)S^2}{\chi ^2 1- \frac{\alpha}{2}}}$}

$\sqrt {\frac{(10-1)4.5^2}{19.02}} < \sigma <\sqrt {\frac{(10-1)4.5^2}{2.70}}$

$3.095 < \sigma < 8.216$

we get the

95 % confidence interval is (3.095, 8.216)

(D)

Here, this 95% cofidence interval contain population standard deviation 4.24 (square root of population variance 18). So, here we can conclude that the The variance is equal to 18 $mg^2$

i.e. The null hypothesis fail to be rejected.