Question

-0.021 0.029 -0.009 -0.002 0.002 -0.006 0.006 -0.064 0.023 0.031 Table 1: A sample of FIR
Your friend seek help from you to use your knowledge from TSTA602 to assist Bank X to do the following data analysis. Please write a report to answer all the following questions.
(a). calculate (mannually) the sample mean, and sample standard de- viation for the sample in Table 1.
(b). if we draw samples of sizes 10 many times and form a distribution of sample mean, state the distribution of the sample mean and provide the reason for your conclusion; calculate the mean of the sample means and its standard deviation.
(c). based on (b), find the probability that the sample mean is smaller than 0.02. Please keep two decimal places in the calculation of standardization.
(d). if the sample mean from (a) is your observed value, calculate the 95% confidence interval for the sample mean.
(e). An outlier is a data point outside the interval [Q1 − 1.5IQR, Q3 + 1.5IQR], where Q1 and Q3 ko are first and third quartile respectively, and IQR
1

is the interquartile range. Explain whether there is any outlier appears in the sample in Table 1? You can calculate Q1 and Q3, and IQR using R (in Rstudio).
(f). Use R (in Rstudio) to generate 1,000,000 samples of size 10 from N(0.032, 0.0016) (please set the seed equals to 602), compute the sample mean for each of these samples, draw a histgram (set the frequency parameter to FALSE) for these sample means and add a density curve to the histgram. Please provide the histgram with the density curve here and attach your R code as appendix.

Answer 1

a.

Sample standard deviation:

b.

According to central limit theorem, distribution of sample mean follows normal distribution with mean and standard deviation .

Hence mean of sampling distribution of sample mean is = - 0.0011 and standard deviation is .

c.

The probability that the sample mean is smaller than 0.02,

d.

95% confidence interval for sample mean:

Degrees of freedom = n - 1 = 10 - 1 = 9

Cut off for 95% confidence interval = t0.05/2, 9 = 2.262

Thus confidence interval is,