CONFIDENCE INTERVAL PROBLEMS

These are two problems that I'm having problems with;

1)

An experimental egg farm is raising chickens to produce low cholesterol eggs. A lab tested 36 randomly selected eggs and found that the mean amount of cholesterol was 200 mg. The sample standard deviation was found to be s = 18.0 mg on this group. Assume that the population is normally distributed.

a. Find the margin of error for a 95% confidence interval. Round your answer to the nearest tenths.
b. Find a 95% confidence interval for the mean m cholesterol content for all experimental eggs. Assume that the population is normally distributed.


2)

The new Twinkle bulb has a mean life of hours with a standard deviation 16 hours. A random sample of 64 light bulbs is selected from inventory. The sample mean was found to be 300 hours.
a. Find the margin of error E for a 90% confidence interval. Round your answer to the nearest hundredths.
b. Construct a 90% confidence interval for the mean life, m of all Twinkle bulbs.

Show a problem you can solve. Give us a clue where you are. Showing no work at all is not helpful.

I can show you what I did...and I really don't think I'm correct..but again, like I said previously, if I knew what I was doing, I would not waste my time by proposing help with the problems that I post. With that being said..this is what I worked out for the two problems that I listed..if someone could take a look at them, and then be so kind to show me step-by-step what I did wrong with them. Thanks so much in advance...

Answer to Problem 1:

Margin of Error:

= 1.96 x 18.0 sqrt 36 = 0.98

Confidence Interval

E = 1.96 - 0.98 = 0.98 E = 1.96 + 0.98 = 2.94

Answer to Problem 2:

Margin of Error:

E = 1.645 x 16 sqrt 64 = 0.42

Confidence Interval

E = 1.645 - 0.42 = 1.225 E = 1.645 + 0.42 = 2.065

We like volunteers! Please provide some clue every time, otherwise we can only guess what might be most helpful. If someone's time must be wasted, how about we don't make it the time of the volunteers.

Looks like we wandered off right away.

z-value = 1.96

Standard Error = 18 / sqrt(36) = 3

Margin of Error = 1.96 * 3 = 5.88

You'll have to show me how you managed 0.98

You seem to have used the value correctly in the construction of the Confidence Interval.

What I did was I multiplied

1.96 x 18.0 divided by sqrt 36 = 0.98

So, how did I do for the second part ? Did I do anything wrong with that one as well? Thanks..I try...

I see. You missed the square root button.

1.98 * 18 / 36 = 0.98

1.98 * 18 / sqrt(36) = 1.98 * 18 / 6 = 5.94

Same problem on the second one.

1.645 * 16 / 64 = 0.42

1.645 * 16 / sqrt(64) = 1.645 * 16 / 8 = 3.29

Okay..so did I do this one correctly: BTW...thanks so much for taking time to review my problems appropriately.

a. Find the margin of error E.
b. Construct a 90% confidence interval for the population proportion p of all Twinkle bulbs.
The new Twinkle bulb is being developed to last more than 1500 hours. A random sample of 100 of these new bulbs is selected from the production line. It was found that 46 lasted more than 1500 hours. Find the point estimate for the population proportion, the margin of error for a 90% confidence interval and then construct to the 90% confidence interval for the population proportion.
Point estimate = (46/100) = 0.46
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a. Find the margin of error E.
E = 1.645*sqrt[0.46*0.36/100] = 0.1656
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b. Construct a 90% confidence interval for the population proportion p of all Twinkle bulbs.
0.46 - 0.1656 < p < 0.46 + 0.1656

What's "0.36"? I was expecting 1 - 0.46 = 0.54.

No worries. We all love to help. We just need a clue, since guessing what is needed is almost never the best way to go.

You are correct ...I transposed 0.36 when it should have gone in another problem that I was solving...Sorry.