Employment Of Sampling Distributions In Statistics - Assignment Solutions

In this assignment you need to use the techniques of Topics 4 - 6 to employ sampling distributions to find probabilities relating to means and proportions, and to make inferences about one and two samples. We estimate that this assignment will take you about six hours to do once you are properly prepared for it.
Rationale Assignment 2 is designed to assess the following learning outcomes:
• be able to use a statistical package to analyse data appropriately, and then interpret the output.
• be able to calculate and interpret probabilities, and use standard probability distributions
• be able to explain the concepts of statistical inference, and apply these to confidence intervals and tests of hypotheses.
• be able to evaluate if the assumptions underlying statistical techniques are valid in a given scenario.
Throughout the assignment you are expected to use R Commander to calculate probabilities,analyse data, perform statistical inference and produceplots.

As per 2016 census, 25% people are living in NSW who have no religion. A random sample of 120 people living in NSW.

Answer 1. a)

            We know that, CDF of binomial distribution is given as-

            F (k; n, p) = Pr (X ≤ k) =

            The probability that no more than 20% of the selected people have no religion =

            Here, n =120, p = 0.25. For, k = 0, 1, 2, …, we find that-

            F (26; 120, 0.25) = Pr (X ≤ 26) = 0.1718 < 0.2.

            Therefore, the probability that no more than 20% of the chosen people have no religion =     

Answer 1. b)

            The conditions that are necessary for the normal approximation to the binomial are-

1) The sample size (n =120) must be sufficiently large. The ‘thumb rule’ shows that normally sample size should be greater than 50 that is satisfied here.

2) The central limit theorem shows that the sum of independent Bernoulli random variables draws conclusions about a true ‘population proportion’ (p).

                      Z =     d     N (0, 1) (Feller 2015)

Answer 1. c)

            The approximated probability using a normal distribution =

Z =  =

            Here,

The hypotheses are-

Null hypothesis (H₀): The average weight of sugar of ‘Diet Coke’ in 25 selected cans is equal to 0 gm.

Alternative hypothesis (H_A): The average weight of sugar of ‘Diet Coke’ in 25 selected cans is greater than 0 gm.

Answer 2. b)

            ‘One Sample t-test’ is used in this context.

Answer 2. c)

            The decision rule is that-

If calculated p-value is greater than level of significance (0.05), then the null hypothesis cannot be rejected. If calculated p-value is lesser than the level of significance, the null hypothesis can be rejected.

Answer 2. e)

            The null hypothesis could be rejected as the calculated p-value (2.2e^-16) is less than the level of significance (5%).

Answer 2. f)

            It could be concluded that the average weight of sugar of ‘Diet Coke’ in 25 selected cans is greater than 0 gm. Therefore, the slogan of Diet Coke ‘No sugar, No Calories’ is proved to be invalid from the test result of sample of 25 randomly selected cans.

Answer 2. g)

            The assumptions for the ‘One sample t-test’ are-

• The observations of the numeric variable are independent of one another.
• The undertaken variable must be continuous in nature.
• The undertaken variable must be normally distributed.
• The variable should not contain any outliers.

            One sample t-test with 8 observations at 5% level of significance shows that-

Answer 2

H₀: μ = 0.

H₁: μ > 0.

            The critical right-tailed t-statistic = 1.859547.

Answer 3. b)

             If a null hypothesis can be rejected at 1% level of significance, then null hypothesis could also be rejected at higher level of significance. As, 10% level of significance is greater than 1% level of significance, therefore, the null hypothesis could be rejected at 10% level of significance also.

Answer 3. c)

            If a null hypothesis cannot be rejected at 5% level of significance, then it cannot be rejected at 1% level of significance. Failure of rejection at 5% level of significance means that there exist 95% evidence for the claim of the basic assertion. It is very much possible that the claim could be fulfilled with 99% evidence. Therefore, the null hypothesis cannot be rejected at 1% level of significance with confirmation.

Answer 3. d)

            The two-sample proportional test at 0.05 level of significance provides the test-statistic = (-1.21). The hypotheses are-

H₀: p₁-p₂ = 0.

H₁: p₁-p₂ ≠ 0.

            The p-value is calculated as 0.226279. The calculated p-value is greater than level of significance (5%). Therefore, the null hypothesis could be rejected with 95% confidence.

Answer 3. e)

            As per the data, it is obtained that 95% confidence intervals for the true mean are (0.53, 0.87). Here, H₀: μ = 0.5 and H₁: μ ≠ 0.5.

            The upper confidence interval,  + Z(_α/2) * S.E.(X) = 0.87 (Steiger and Fouladi 2016).

            The lower confidence interval,  - Z(_α/2) * S.E.(X) = 0.53.

The calculated mean () = 0.7

            Here, Z(_1-α/2) = 1.96, where α =0.05. Therefore, S.E.(X) =

            The calculated Z-statistic =

            The p-value found out from the calculated Z-statistic = 0.021119. It is less than the level of significance (5%). Therefore, the null hypothesis could be rejected with 95% probability.

Answer 3. f)

            Welch’s two sample t-test is used for testing equality of means of two independent samples preferably of unequal variances and unequal sample sizes. However, the equality of average weights of experimented 10 rats in prior-treatment and post-treatment situations would be preferably tested by the paired t-test. Here, two variables occur from same population, from two different situations and of equal sizes.

            Therefore, paired t-test would be more preferably used rather than Welch’s two sample t-test.      

            The test statistic to estimate the proportion of all families that would find the program effective is Z-statistic.

Answer 3

            The calculated value of Z-statistic =  =

            Note that, here,

Answer 4. b)

            The hypotheses are-

H₀:

H₁:

            The calculated one-tailed p-value of the Z-statistic (0.876) = 0.190515. Therefore, the null hypothesis cannot be rejected with 95% confidence. Hence, the company’s claim of program ‘Sleeping well’ could be retained supported by data.

Answer 4. c)

            The margin of error for 95% confidence interval is not greater than 0.03.

            M.E. = Z *  = 0.03.

            Or,

            Or,

            Or,  = 1060.

            The number of families needed to construct a 95% confidence interval for a similar trial so that the margin of error is not greater than 0.03 is 1060.

Answer 4. d)

            The ‘Sleeping well’ has proportion =  =

            The ‘Sleeping well 2’ has proportion =  =

            The pooled proportion of families are

            The calculated pooled proportion of families = 0.0014716702.

Answer 4. e)

             We set the hypothesis-

            H₀: The effectiveness of ‘Sleeping well 2’ () is equal to the effectiveness of ‘Sleeping well’ (.

            H_A: The effectiveness of ‘Sleeping well 2’ () is higher than the effectiveness of ‘Sleeping well’ (.

            The calculated t-statistic =  = 28.2186 (De Winter 2013).

            Degrees of freedom = (

            The calculated p-value of T (28.2186, 318) = 0.00001.

            As the calculated p-value (0.00001) is less than 0.05, therefore the null hypothesis of equality of means of both the proportions could be rejected. The alternative hypothesis can not be rejected in this context. It could be interpreted that effectiveness of ‘Sleeping well 2’ is higher than the effectiveness of ‘Sleeping well’. The hypothesis testing indicates the improvement of effectiveness.

            Null Hypothesis (H₀): The average sugar content of Coke Zero and Diet Coke does not differ.

            Alternative Hypothesis (H_A): The average sugar content of Coke Zero and Diet Coke differ.

            The two-sample t-test is applied to determine the difference of average sugar content of Coke Zero and Diet Coke. The calculated t-statistic (0.95939, d.f. = 24) generates the p-value = 0.3469. The calculated p-value is greater than 0.1 (assumed level of significance).

            Hence, the null hypothesis cannot be rejected with 90% confidence. The difference of average sugar level of two kinds of Cokes is significant. Therefore, it is 90% evident that the average sugar level of Coke Zero is equal to the average sugar level of Diet Coke.

Answer 5. b)

            The statistical assumptions of the two-sample t-test are-

• Both populations must be normal in nature.
• The standard deviation of both populations must be equal. Hence, the variance would be homogeneous.
• Both samples have to be randomly drawn independent of each other.
• Two samples must be reasonably large in size.

The assumption of normality of both the variables is right. These are also selected randomly and these are independent to each other. However, the standard deviations of both variables are not equal. Also, the sample sizes are not adequately large (n = 25). Hence, some assumptions are maintained here and some are violated.

Answer 5. c)

            Actually, a paired sample t-test compares two sample means from the same population of same variable whether the means are different in two various times or not (pre-test and post-test). On the other hand, two-sample t-test compares means from different populations whose members have been matched for determining whether the difference between two means of two variables is equal to 0 or not.

            It could be interpreted that as ‘Coke Zero’ and ‘Diet Coke’ are from two different populations, therefore, two-sample t-test is more preferable than paired-sample t-test.  

References:

De Winter, J.C., 2013. Using the Student's t-test with extremely small sample sizes. Practical Assessment, Research & Evaluation, 18(10).

Feller, W., 2015. On the normal approximation to the binomial distribution. In Selected Papers I (pp. 655-665). Springer, Cham.

Steiger, J.H. and Fouladi, R.T., 2016. Noncentrality interval estimation and the evaluation of statistical models. What if there were no significance tests, pp.197-229.