Question

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution. A B [26,30) 7 3 [30,34) 12 9 [34,38) 15 19 [38,42) 7 10 Total 41 41 Compute a 95% CI for the difference between the true proportions of all specimens of alloys A and B that have an ultimate strength of at least 34 ksi.

Answer 1

The 95% confidence interval would be given (-0.377;0.0366).

We are confident at 95% that the difference between the two proportions is between
`-0.377 \leq p_B -p_A \leq 0.0366`

Explanation:

The data given is:

A B

________________________________

[26,30) 7 3

[30,34) 12 9

[34,38) 15 19

[38,42) 7 10

________________________________

Total 41 41

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`p_A` represent the real population proportion for brand A

`\hat p_A =(15+7)/(41)=0.537` represent the estimated proportion for Brand A

`n_A=41` is the sample size required for Brand A

`p_B` represent the real population proportion for brand b

`\hat p_B =(19+10)/(41)=0.707` represent the estimated proportion for Brand B

`n_B=41` is the sample size required for Brand B

`z` represent the critical value for the margin of error

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

The confidence interval for the difference of two proportions would be given by this formula

`(\hat p_A -\hat p_B) \pm z_(\alpha/2) \sqrt{(\hat p_A(1-\hat p_A))/(n_A) +(\hat p_B (1-\hat p_B))/(n_B)}`

For the 95% confidence interval the value of
`\alpha=1-0.95=0.05` and
`\alpha/2=0.025`, with that value we can find the quantile required for the interval in the normal standard distribution.

`z_(\alpha/2)=1.96`

And replacing into the confidence interval formula we got:

`(0.537-0.707) - 1.96 \sqrt{(0.537(1-0.537))/(41) +(0.707(1-0.707))/(41)}=-0.377`

`(0.537-0.707) + 1.96 \sqrt{(0.537(1-0.537))/(41) +(0.707(1-0.707))/(41)}=0.0366`

And the 95% confidence interval would be given (-0.377;0.0366).

We are confident at 95% that the difference between the two proportions is between
`-0.377 \leq p_B -p_A \leq 0.0366`