SOLUTION: Hello, Iam having some trouble with my statistics homework, please I hope some one can help me.I have two questions: : Confidence Intervals for the Mean (Small Samples) 1) A

Question 186258: Hello, Iam having some trouble with my statistics homework, please I hope some one can help me.I have two questions:
: Confidence Intervals for the Mean (Small Samples)
1) An experimental egg farm is raising chickens to produce low cholesterol eggs. A lab tested 25 randomly selected eggs and found that the mean amount of cholesterol was 190 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  cholesterol content for all experimental eggs. Assume that the population is normally distributed.

Confidence Intervals for Population Proportions
2) The new Twinkle bulb is being developed to last more than 1000 hours. A random sample of 100 of these new bulbs is selected from the production line. It was found that 75 lasted more than 1000 hours. Find the point estimate for the population proportion, the margin of error for a 95% confidence interval and then construct to the 95% confidence interval for the population proportion p.

a. Find the margin of error E.

b.Construct a 95% confidence interval for the population proportion p of all Twinkle bulbs.

I would like to thank you in advance for all your help and using your time to help me.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Confidence Intervals for the Mean (Small Samples)
1) An experimental egg farm is raising chickens to produce low cholesterol eggs. A lab tested 25 randomly selected eggs and found that the mean amount of cholesterol was 190 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.
----
sample mean: 190 mg
E = t = 2.064 = 2.064= 7.4304
---------
95% CI: 190-7.4304 < u < 190+7.4304
--------------------------------------------------

b.Find a 95% confidence interval for the mean  cholesterol content for all experimental eggs. Assume that the population is normally distributed.
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Some word(s) is missing from your post.
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Confidence Intervals for Population Proportions
2) The new Twinkle bulb is being developed to last more than 1000 hours. A random sample of 100 of these new bulbs is selected from the production line. It was found that 75 lasted more than 1000 hours. Find the point estimate for the population proportion, the margin of error for a 95% confidence interval and then construct to the 95% confidence interval for the population proportion p.
----
point estimate: 75/100 = 0.75
E = z = 1.96[sqrt(0.75*0.25/100] = 0.0849
--------------------
95% CI: 0.75-0.0849 < p < 0.75+0.0849
==========================================================
Cheers,
Stan H.
a. Find the margin of error E.
b.Construct a 95% confidence interval for the population proportion p of all Twinkle bulbs.

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