
Somebody please please help with these questions!! 10 points straight away for right answer!?
Question 7
The amount of silver in a Roman denarius coin is Normally distributed with standard deviation σ = 0.3 grammes. A sample of 15 denarii is weighed, and it is discovered that the sample mean weight of silver is 3.87 grammes. Calculate the upper end of the 90% confidence interval for the mean weight of silver in a denarius based on these data, giving your answer to the nearest 0.01 grammes.
Question 8
A sample of 17 observations is taken from a Normal distribution with unknown mean μ and unknown variance σ2. The sample mean is 82.1, the sample variance 57.6. Calculate the lower end of the 99% confidence interval for μ based on these data.
7) If the standard deviation is 0.3 and the sample size is 15, then the standard deviation of the sample mean is σ/√n = 0.3/√15. So the confidence interval is:
3.87± (Z)(0.3/√15), where “Z” is the Z value for a confidence level of 90%. This should be the Z value for (1-0.9)/2. Look this up in your Z-table, and you should get -1.6449. You’re asked to give the upper end of the interval, so just cacluate the upper end.
You’ve got the sample mean and sample variance, so use the same formula for the interval (sample mean ± Z*[sample var.]), but look up the Z value for (1-0.99)/2 and take the lower end.
If Roman Denarius $64, Silver $530 / oz, $38 / Silver Dime.
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