Question

In an experiment involving the breaking strength of a certain type of thread used in personal flotation devices, one batch of thread was subjected to a heat treatment for 60 seconds and another batch was treated for 120 seconds. The breaking strengths (in N) of ten threads in each batch were measured. The results were 60 seconds: 43 52 52 58 49 52 41 52 56 51 120 seconds: 59 55 59 66 62 55 57 66 66 51 Let μX represent the population mean for threads treated for 120 seconds and let μY represent the population mean for threads treated for 60 seconds. Find a 99% confidence interval for the difference μX−μY . Round down the degrees of freedom to the nearest integer and round the answers to three decimal places. The 99% confidence interval is ( , ).

Answer 1

The 99% confidence interval is (-20.774, 2.774)

Explanation:

The 60 seconds and 120 seconds breaking strength measures are;

60 seconds: 43, 52, 52, 58, 49, 52, 41, 52, 56, 51

120 seconds: 59, 55, 59, 66, 62, 55, 57, 66, 66, 51

The Mean and standard deviation for the data are obtained as follows;

The mean strength for the 60 seconds treatment samples, μX = 50.6 N

The standard deviation for the 60 seconds treatment samples, sX = 5.211099 N

The number of threads in the sample, n₁ = 10

The mean strength for the 60 seconds treatment samples, μY = 59.6 N

The standard deviation for the 60 seconds treatment samples, sY = 5.295701 N

The number of threads in the sample, n₂ = 10

The 99% confidence interval for the difference in mean, is given as follows;

`\left (\mu X- \mu Y \right )\pm t_{\left((\alpha)/(2), df \right) } \cdot \sqrt{(sX^(2))/(n_(1))+(sY^(2))/(n_(2))}`

The degrees of freedom, df = n₁ + n₂ - 2 = 20 - 2 = 18

(1 - α)·100 = 99%

α = 1 - 99/100 = 0.01

α/2 = 0.01/2 = 0.005

The critical-t at 0.005 significant level and df = 18 is 2.878

The confidence interval is therefore;

`C.I. = \left (50.6- 59.6 \right )\pm 2.878 \right) } * \sqrt{(5.211099^(2))/(10)+(5.295701^(2))/(10)}`

C.I. = -9 ± 11.774425

∴ By rounding to three decimal places, the 99% confidence interval is (-20.774, 2.774).