Question

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SSCE 2193 c) A company produces metal pipes of a standard length, and claims that the standard deviation of the length is at most 1.1 cm. One of its clients decides to test this claim by taking a sample of 25 pipes and checking their lengths. They found that the standard deviation of the sample is 1.5 cm. Can we accept the company's claim at 2.5% significance level? (5 marks) QUESTION 6 (16 MARKS) a) Three different arrangements of instruments on a control panel of an airplane were tested by simulating an emergency condition and observing the reaction time required to correct the condition. The reaction time (in tenths of a second) of 21 pilots (randomly assigned to the different arrangements) were as follows: 4 SSCE 2193 Arrangement 3 Arrangement 2 10 Arrangement 1 12 13 10 15 13 12 Can we statistically conclude at 1% significance level that the different arrangement has affected the reaction time? Use one way ANOVA (10 marks) QUESTION 7 (12 MARKS) The following data shows the drying time of a certain varnish and the amount of an additive that is used to reduce the drying time. Amount of varnish additive (grams) Drying time (in hours) 0 12.0 10.5 10.0 4 7.0 7.5 8.5 9.0 Find the least square estimated regression line of the drying time (in hours) against the amount of varnish additive (grams) (2 marks) ii. Estimate the drying time of the varnish when 6.5 grams of the additive is being used (1 mark) iii. Can we conclude that there exist a negative relationship between the drying time of a certain varnish and the amount of an additive at 5% significance level? (7 marks) iv. Use Pearson product moment correlation coefficient, to measure the strength and direction of a linear relationship between the amount of varnish additive (grams) and the drying time (in hours). Comment on the answer (2 marks)

Answer 1

C) We will set up the null hypothesis

Ho : σ2 = σ。

1:ơ

under null hypothesis the test statistics is

2 (n 1)s2 O02

we know that
n-25
,
(1.1)2-121 6
and
s2 = (1.5)2-225

(25 - 1) 2.25 2 1.21

2 = 44.62

Also $\chi^2$ tabulated at 2.5% of significance level with 24 degree of freedom = 39.36.

Since calculated $\chi^2$ is greater then tabulated $\chi^2$ at 2.5% level of significance Hence we will accept our null hypothesis and concludes that true standard deviation = 1.1.

Question 6 a) We will set up null hypothesis that

$H_{0}:$ All arrangement are same.

$H_{1}:$ At least two arrangement are different.

Grand mean is obtained as follow

R n 71

1413+9+ 8+8) - 10 21

Total sum of square are obtained as

k n

$TSS=(14-10)^2+(13-10)^2+...+(8-10)^2+(8-10)^2=136.00$

Treatment sum of square is obtained as follow

$SSTR=\sum_{i=1}^{k}n_{i}\bar{yi.}^2-n\bar{y}^2$ where $\bar{yi.} =\sum_{j=1}^{ni} \frac{\bar{yi.}}{ni}$ are treatment means

Treatment means are given below in the table.

Arrangement 1	Arrangement 2	Arrangement 3
14	10	11
13	12	5
9	9	9
15	7	10
11	11	6
13	8	8
	12	8
	9
$\bar{y1.}$ = 12.5	$\bar{y2.}$ = 9.75	$\bar{y3.}$ = 8.143

$SSTR=(6*12.5^2+8*9.75^2+7*8.143^2)-21*10^2$

$SSTR=62.14$

We also know that

$SSE=TSS-SSTR=136.00-62.14=73.86$

The complete ANOVA table is given below

Source	DF	SS	MS	F	P.Value
Treatment	2	62.14	31.07	7.57	0.004
Error	18	73.86	4.1
Total	20	136

Since calculated P.Value is less then 0.01 hence we will reject null hypothesis and conclude that At least two arrangement are different.

Question 7) The regression estimate are obtained with the following formula.

$b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}$,  $a=\frac{\sum y- b\sum x}{n}$

The calculation are given below

X	Y	X^2	Y^2	XY
0	12.0	0	144	0
1	10.5	1	110.25	10.5
2	10.0	4	100	20
3	8.0	9	64	24
4	7.0	16	49	28
5	8.0	25	64	40
6	7.5	36	56.25	45
7	8.5	49	72.25	59.5
8	9.0	64	81	72
36	80.5	204	740.75	299

$\therefore b=\frac{9*299-36*80.5}{9*204-(36)^2}=- 0.383$

$a=\frac{80.5- (- 0.383)*36}{9}=10.5$

i) The least square regression is

Y = 10.5 - 0.383 X

ii) for 6.5 grams of varnish we have

Y = 10.5 - 0.383 * 6.5

Y = 8.01

iii)we will set up null hypothesis

under null the test statistics is

$t=\frac{b}{S_{ b}}\sim t_{n-2}$

$S_{ b}=\frac{S.E_{Residual}}{\sqrt{\sum X^2-\frac{(\sum X)^2}{n}}}$ where $S.E_{Residual}= \sqrt{\frac{\sum(Y-\hat{Y})^2}{n-2}}$

The table is given below

Y	$\hat{Y}$	(Y -Y)2
12.0	10.5	2.25
10.5	10.117	0.15
10.0	9.734	0.07
8.0	9.351	1.83
7.0	8.968	3.87
8.0	8.585	0.34
7.5	8.202	0.49
8.5	7.819	0.46
9.0	7.436	2.45
80.5	80.712	11.91

$S.E_{Residual}= \sqrt{\frac{11.91}{9-2}}=1.304$

$S_{ b}=\frac{1.304}{\sqrt{204-\frac{(36)^2}{9}}}=0.1684$

$t=\frac{-0.383}{0.1684}=-2.28$

and its corresponding P.value = 0.057 which is greater the 0.05. Hence we will accept null.

Iv) pearson correlation is obtained with following formula

$r = \frac{n\sum xy-\sum x \sum y}{\sqrt{n\sum x^{2}-(\sum x)^{2}} \sqrt{n\sum y^{2}-(\sum y)^{2}}}$