Question

(b) Using the data provided, in Excel, find the sample average and the sample standard deviation. Find the sample size, find the degrees of freedom, find the t-critical value.

(c) Compute a 95% confidence interval for the population average. State your confidence interval in words, clearly stating the context, and also show the answer in the ...< µ < ... format.

(d) Using Excel, given a production run of size N=200 and the sample size of n=30, compute the number of different samples that could be taken of size n=30. Answer with and without replacement.

(e) With respect to the above answer, explain what is meant by “95%” in the 95% confidence interval.

(f) Does the 95% confidence interval cover 95% of the Raisinets sample data? What do you conclude?

Raisinet's Data

Exercises: Section 1.      NetWt 46.9 47.1 45.7 45.6 45.1 48.9 46.5 47.9 48 45.9 46.9 46.2 43.8 45.3 46.2 45.9 46.5 46.7 46.5 46.8 47.1 43.4 46.2 47.3 47.2 45.2 47 46.4 44.3 47.1

Answer 1

b) The sample average is

$\overline{x} = \frac{\sum X_{i}}{30} = \frac{1389.6}{30} = 46.32$

The sample standard deviation is

$s = \sqrt{\frac{\sum(X_{i} - \overline{x})^2}{n-1}}= 1.192$

The sample size is = 30

The degree of freedom is = n- 1 = 30 - 1 = 29

The t-critical is = 2.045

c) The 95% confidence interval for the population average is

$= \overline{x} \pm t_{\alpha/2, n-1} * \frac{s}{\sqrt{n}}$

$= 46.32 \pm 2.045 * \frac{1.192}{\sqrt{30}}$

There is a 95% chance that the population average will lie in the interval 45.875 - 46.765.

The 95% confidence interval for the population average is

$45.875 < \mu < 46.765$

d) Number of samples of 30 taken without replacement is

$= \binom{200}{30}$

Number of samples of 30 taken with replacement is

e) Based on the above question, if we would take all the samples of size 30 and calculate the average value for each of those samples then approximately 95% of those sample means will lie in the interval that we have created above.

f) Number of values from the data which lie in the confidence interval 45.875 - 46.765 is = 10

The percentage of values out of 30 which lies in the 95% confidence interval is = 10/30 * 100 = 33.33%

The 95% confidence interval doesn't contain 95% of the sample data.

Conclusion: The 95% confidence interval doesn't contain 95% of the sample data because the confidence interval is for the average value and not the individual values. 95% confidence interval will contain 95% of the sample means and not 95% of the values of each sample.

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