Calculating Variance and Standard Deviation in Sample

 

Calculating Variance and Standard Deviation in Sample 

A.

a. To compute the mean of the population, you sum up all the values and divide by the number of values:

Population Mean = (8 + 14 + 16 + 10 + 11) / 5 = 59 / 5 = 11.8

b. To select a random sample of size 2 from the population, you can use the `sample` function in R:

```

population <- c(8, 14, 16, 10, 11)

sample_size <- 2

sample_data <- sample(population, size = sample_size)

```

The sample in this case can be (14, 11)

c. To compute the mean and standard deviation of your sample, the following code can be used:

```

sample_mean <- mean(sample_data)

sample_std_dev <- sd(sample_data)

```

For the sample (14, 11), the sample mean is 12.5, and the sample standard deviation is approximately 2.12.

d. Population Mean: 11.8 (calculated in part a)

Sample Mean: 12.5 (calculated in part c)

Population Standard Deviation: Approximately 2.24 (calculated using sd(population))

Sample Standard Deviation: Approximately 2.12 (calculated in part c)

In this case, the sample mean is slightly higher than the population mean, and the sample standard deviation is slightly lower than the population standard deviation. 

B.

1. Does the sample proportion p have approximately a normal distribution?

   - The sample proportion p, which is based on a sample size n = 100 and population proportion p = 0.95, can be assumed to have an approximately normal distribution if both np and nq are greater than or equal to 5. In this case:

   np = 100 * 0.95 = 95

   nq = 100 * (1 - 0.95) = 5

   Since both np and nq are greater than or equal to 5, the sample proportion p is expected to have an approximately normal distribution.

2. What is the smallest value of n for which the sampling distribution of p is approximately normal?

   - The rule of thumb is that for a sample proportion to have an approximately normal distribution, both np and nq should be greater than or equal to 5. Therefore, the smallest value of n for which the sampling distribution of p is approximately normal is when:

   np ≥ 5

   nq ≥ 5

   Solve for n:

   n * 0.95 ≥ 5

   n * (1 - 0.95) ≥ 5

   n ≥ 5 / 0.95 ≈ 5.26 (rounded up to the nearest whole number)

   So, the smallest value of n for which the sampling distribution of p is approximately normal is 6.