7. In a sample of 100 steel canisters, the mean wall...

7. In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. a. Find a 95% confidence interval for the mean wall thickness of this type of canister. b. Find a 99% confidence interval for the mean wall thickness of this type of canister. c. An engineer claims that the mean thickness is between 8.02 and 8.18 mm. With what level of confidence can this statement be made? d. How many canisters must be sampled so that a 95% confidence interval will specify the mean wall thickness to within ±0.08 mm? e. How many canisters must be sampled so that a 99% confidence interval will specify the mean wall thickness to within ±0.08 mm?

Transcript

00:01 So we have a sample size of 100, and the x bar from that was 8 .1 millimeters with a standard deviation of 0 .5 millimeters.

00:12 And in part a, we want to find a 95 % confidence interval for the mean.

00:19 So we're going to take our 8 .1 plus or minus, and then i used my inverse t, my inverse t, and i'm plugged in the area of 0 .0 .0 .0 .0.

00:31 0 .025, there'd be 0 .025 in the lower tail, and then 99 degrees of freedom.

00:37 And i got the t value for that setting, because my table, i could get very close by looking up 100 degrees of freedom, but technically up 99 degrees of freedom.

00:48 And that value came out to be 1 .984.

00:51 And then we take our sample standard deviation divided by the square root of 100 or divided by 10.

00:57 And so 8 .1 minus 1 .1 .1 .1.

01:01 Point nine eight four times point five divided by ten gives me my lower limit to be eight point zero zero zero eight to an upper limit of changing that subtraction sign into an addition sign gives me to eight point one nine nine nine two so there's our ninety five percent confidence interval and now the only thing we want to differently is find a 99 % confidence interval for the mean and again that means i used my inverse t putting 0 .005 in the lower tail with 99 degrees of freedom and that changes this value to 2 .626 so i'm just going to go back and change those values and change that number to 2 .626, and it will be wider naturally.

02:08 And so that goes from 7 .9687 to changing that back to an addition sign to 8 .2313.

02:23 Now, part c, we have an engineer that says that the interval is from 8 .02 to 8 .18.

02:35 Millimeters and we want to find what is that confidence level and so we can see that the margin of air we have a t value times 0 .5 and then divided by the square root of 100 which is 10 and we know that that margin of air the mean is 8 .1 so that margin of air is 0 .08 and when we do our calculation there and multiply by 10 and divide by 0 .02 we get that t value is equal to 1 .6 so we're dealing with a confidence interval that uses a positive 1 .6 for a t value with 99 degrees of freedom and a negative 1 .6 for that value.

03:18 And we want to know what that area is between those.

03:21 So we can find that by using our t cdf and have it go from negative 1 .6 to 1 .6 and have 99 degrees of freedom.

03:34 And that tells us that it's approximately 89 % confidence...

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