Question

You are the construction manager of a company that build standard shopping centres across the world. You meet a supplier who promises you material that allows builders to build significantly quicker than what the existing material allows, without compromising the quality, safety etc of the structures. The CEo asks you to assess during the year and give a report at the end of the year with a recommendation of whether to buy the new material or not. You take a sample of projects from around the world and use new materials from some and old materials from others. This is the information gathered:

Material	New (experiment)	Existing/Old
Sample size of projects	nx = 12	ny = 8
Avg time taken to complete a building	x = 6 months	y = 8 months
Standard deviation	sx = 3 months	sy =2 months

Use the p value method to determine if the suppliers claim is valid to decide whether to continue using the existing material or adopt the new method post the 1 year of experimentation.

1) Decision criteria

Ho

H1

You need to perform a _____ tailed test

2) Assumption:

3) Degrees of freedom

4) Choice of distribution and motivation

5) Test statistic (show formula and all workings)

6) P-value (all workings and analysis of p-value)

7) Decision

8) Interpretation and conclusion

9) what would your decison be if

p-value < x

p-value > x

Answer 1

n1 = 12

= 6

s1 = 3

n2 = 8

= 8

s2 = 2

1) The null and alternative hypothesis is

We need to perform the 2 sample t (independent) test.

For doing this test first we have to check the two groups have population variances are equal or not.

The null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

F = 9 / 4 = 2.25

Degrees of freedom => n1 - 1 , n2 - 1 => 12 - 1 , 8 - 1 => 11 , 7

Critical value = 3.603 ( Using f table)

Critical value > test statistic so we fail to reject null hypothesis.

Conclusion: The population variances are equal.

So we have to use here pooled variance.

2) Assumption: The population variances are equal and population distribution follows the normal distribution.

3) Degrees of freedom = n1 + n2 - 2 = 12 + 8 - 2 = 18

4) T distritution.

5) Test statistic is

Degrees of freedom = n1 + n2 - 2 = 12 + 8 - 2 = 18

6) P-value = 2*P(T< - 1.65) = 2*0.0581 = 0.1163

7) P-value > 0.05 we fail to reject the null hypothesis.

8) Conclusion: There is sufficient evidence to support the supplier's claim is valid to decide whether to continue using the existing material or adopt the new method post the 1 year of experimentation.