Question

1) A medical researcher wants to investigate the amount of time it takes for patients' headache pain to be relieved after taking a new prescription painkiller. She plans to use statistical methods to estimate the mean of the population of relief times. She believes that the population is normally distributed with a standard deviation of 22 minutes. How large a sample should she take to estimate the mean time to within 4 minutes with 96% confidence?

Sample Size =

2) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.88 million dollars. Assuming a population standard deviation gross earnings of 0.48 million dollars, obtain a 99% confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions).
Confidence interval: ( , )

b) Which of the following is the correct interpretation for your answer in part (a)?
A. We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the interval
B. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval
C. There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the interval
D. None of the above

3. Use the given data to find the 95% confidence interval estimate of the population mean μμ. Assume that the population has a normal distribution. Give your answers to 2 decimal places.

IQ scores of professional athletes:
Sample size n=15
Mean x =105
Standard deviation s=12

equation editor

-________<μ<__________

Answer 1

1) The minimum sample is calcultaed as

Where, E = Margin of error,

And Z value is calculated using Z table shown below.

2)

The formula for estimation is:

μ = M ± Z(s_M)

where:

M = sample mean
Z = Z statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 2.88
Z = 2.58
s_M = √(0.48²/30) = 0.09

μ = M ± Z(s_M)
μ = 2.88 ± 2.58*0.09
μ = 2.88 ± 0.2257

99% CI [2.6543, 3.1057].

b) Interpretation:

B. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the interval

3) Here Population standard deviation is unknown and sample size is also less han 30 so, t-distribution is applicable here.

he formula for estimation is:

μ = M ± t(s_M)

where:

M = sample mean
t = t statistic determined by confidence level
s_M = standard error = √(s²/n)

M = 105
t = 2.14
s_M = √(12²/15) = 3.1

μ = M ± t(s_M)
μ = 105 ± 2.14*3.1
μ = 105 ± 6.65

5% CI [98.35, 111.65].

The Z Table

T -Table for T values