Question

A political candidate has asked you to conduct a poll to determine what percentage of people support her.

If the candidate only wants a 4% margin of error at a 95% confidence level, what size of the sample is needed?

The political candidate will need to sample people.

You measure 39 textbooks' weights, and find they have a mean weight of 44 ounces. Assume the population standard deviation is 12.7 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight.

Give your answers as decimals, to two places

I am 99% confident that the mean weight of textbooks is between and ounces.

You measure 46 textbooks' weights and find they have a mean weight of 38 ounces. Assume the population standard deviation is 4.8 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight.
A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 30 bacteria reveals a sample mean of ¯x=80x¯=80 hours with a standard deviation of s=5s=5 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.5 hours at a 95% level of confidence.

What sample size should you gather to achieve a 0.5 hour margin of error?

He would need to sample bacteria.
Give your answers as decimals, to two places

I am 99% confident that the mean weight of textbooks is between and ounces.

Answer 1

1) Let the true proportion of people who support her be P.

Margin of errror = E = 4% = 0.04

Confidence level = 95% = 0.95

Therefore, the level of significance = 1 - 0.95 = 0.05

Now, formula for finding the sample size is -

Since, level of significance = = 0.05,

where, is the critical value of z for two tailed test at 0.05 level of significance = 1.96

E = margin of errror = 0.04

p is the prior estimate of true proportion of people who support her. Since, we don't have any prior estimate for p, we will take p = 0.5, as for p = 0.5, the product p(1-p) will have highest value, and this will provide us the greater sample size, hence, high accuracy.

So, p = 0.5

Putting all these values in the sample size formula, we get, sample size as -

~ 600

2) Sample size = n = 39

Sample mean weight = = 44 ounces

Population standard deviation = = 12.7 ounces.

Confidence level = 99% = 0.99

Level of significance = = 1 - 0.99 = 0.01

99% confidence interval for the true population mean textbook weight is given by -

is the critical value of z for two tailed test at 0.01 level of significance = 2.58

  

= [38.75, 49.25]