Question

One question in the survey asked how much time per year the children spent in volunteer activities. The sample mean was 14.76 hours and the sample standard deviation was 16.54 hours.

Required:

a. Based on the reported sample mean and sample standard deviation, explain why it is not reasonable to think that the distribution of volunteer times for the population of South Korean middle school students is approximately normal.
b. The sample size was not given in the paper, but the sample size was described as large. Suppose that the sample size was 500. Explain why it is reasonable to use a one-sample t confidence interval to estimate the population mean even though the population distribution is not approximately normal.
c. Calculate and interpret a confidence interval for the mean number of hours spent in volunteer activities per year for South Korean middle school children.

Answer 1

a. If the distribution was normal, many values would be negative, what is incompatible with the response variable (hours dedicated to volunteer activities).

b. If the sample is big, accordingly to the Central Limit Theorem, the sampling distribution shape tends to be normally-like, so we can apply a one-sample t-test.

c. The 95% confidence interval for the mean is (13.307, 16.213).

Explanation:

a. If the distribution was normal, the values with one or more standard deviation below the mean would be negative, what is incoherent for this case. This, in a normal distribution, represents approximately 16% of the values.

If we calculate the probabilty for a normal distribution with the sample parameters, the probability of having "negative hours" is 18.6% (see picture attached).

b. If the sample is big, accordingly to the Central Limit Theorem, the sampling distribution shape tends to be normally-like, so we can apply a one-sample t-test.

The sampling distribution standard deviation is also reduced by a factor of 1/√n.

c. We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=14.76.

The sample size is N=500.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

`s_M=(s)/(√(N))=(16.54)/(√(500))=(16.54)/(22.3607)=0.7397`

The t-value for a 95% confidence interval is t=1.965.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=1.965 \cdot 0.7397=1.453`

Then, the lower and upper bounds of the confidence interval are:

`LL=M-t \cdot s_M = 14.76-1.453=13.307\\\\UL=M+t \cdot s_M = 14.76+1.453=16.213`

The 95% confidence interval for the mean is (13.307, 16.213).