Explanation:
First, let's define some variables:
μ = population mean
σ = population standard deviation
p = population proportion
s = standard error (sample standard deviation)
n = sample size
x = sample mean or proportion
Now, our equations.
Given the population deviation, the standard error can be calculated as:
s = σ / √n
For a proportion, the standard error is calculated as:
s = √(p (1 − p) / n)
If the population is normally distributed, or if the sample size is greater than 30, you can use the z score to find the probability:
z = (x − μ) / σ
Confidence intervals can be constructed using:
x = (μ or p) ± (t or z) × s
The product of the critical value (t or z) and the standard error is also known as the margin of error.
4. Find the standard error:
s = σ / √n
s = 3 / √60
s = 0.387
Since the population is normally distributed, the sample is also normally distributed. Finding the z score:
z = (x − μ) / s
z = (9.5 − 10) / 0.387
z = -1.29
Using a z-score table, the probability is therefore:
P = 1 − 0.0985
P = 0.9015
5. Find the standard error:
s = √(p (1 − p) / n)
s = √(0.65 × (1 − 0.65) / 180)
s = 0.0356
Since the sample size is greater than 30, it can be approximated as a normal distribution. The z-score is:
z = (0.70 − 0.65) / 0.0356
z = 1.41
Since this is not more than 2 standard deviations, the statistic is not significant (we should not be surprised if 70% of the seeds germinate).
6. First, find the proportion.
p = 29 / 634
p = 0.0457
Find the standard error:
s = √(p (1 − p) / n)
s = √(0.0457 × (1 − 0.0457) / 634)
s = 0.0083
Since the sample size is greater than 30, we can use the z-score as the critical value. At 99% confidence, z = 2.576. Therefore, the confidence interval is:
x = 0.0457 ± 2.576 × 0.0083
x = (0.0214, 0.0671)
We are 99% confident that the percent of all deaths that are caused by accidents is between 2.14% and 6.71%.
7. Find the standard error.
s = 15.7 / √25
s = 3.14
Find the critical value. Since the sample size is less than 30, we'll need to use t score. For 24 degrees of freedom and 90% confidence, t = 1.711. Therefore, the confidence interval is:
x = 225 ± 1.711 × 3.14
x = (219.6, 230.4)
We are 90% confident that the mean cholesterol content of all such eggs is between 219.6 mg and 230.4 mg.
8. Let's assume that the needed sample size is greater than 30, so we can use the z-score. At 99% confidence, z = 2.576. The margin of error is 4, so the standard error is:
4 = 2.576 × s
s = 1.553
So the sample size is:
1.553 = 12 / √n
n = 60
9. Let's assume that the needed sample size is greater than 30, so we can use the z-score. At 95% confidence, z = 1.960. The margin of error is 0.04, so the standard error is:
0.04 = 1.960 × s
s = 0.020
Since we don't know the value of p, we set p = 0.5 to maximize the sample size needed. So the sample size is:
0.020 = √(0.5 (1 − 0.5) / n)
n = 600