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Unit 14 Analysis of Quantitative Data: Inferential Statistics Based on Parametric Tests

1. List the assumptions of parametric tests.

a) Variables described are expressed in interval or ratio scales.
b) Samples have equal or nearly equal variances.
c) Observations are independent.

2. Compute: (i) .95 and (ii) .99 confidence intervals for the population mean for a sample with mean = 30, SD = 5 and N = 100.

29.02 and 30.98 

28.71 and 31 .29 

3. Find the .99 confidence interval for the population mean for a sample of 5 students whose scores on a test are 6, 5, 7, 4, 5. 

3.05 and 7.70

4. The means and standard deviations of two independent large samples for 95 nuclear and 64 joint families on an attitude test are as follows:

| Family Type | N | Mean | Standard Deviation 

| Nuclear | 95 | 70 | 8 |

| Joint | 64 | 65 | 9 |

Test the hypothesis that the two samples come from the same population at .05 and .01 level.

a) The calculated t-value is 3.67
b) The means of the two samples are significantly different at the .05 and .01 levels. The null hypothesis is rejected.

5. Explain the concept of one-tailed and two-tailed tests.

In a one-tailed test, the hypothesis specifies the direction of the difference. A two-tailed test does not specify the direction of the difference.

6. In a study on the effectiveness of counselling, a group of clients are pre-tested, given counselling and then post-tested. The mean gain score was 10 with standard deviation as 1. The value of t is 10. The investigator used a one-tailed test. Test the null hypothesis at .05 and .01 levels of significance at 9 degree of freedom (df).

At df of 9, the .05 significance level is 1.83 and the .01 level is 2.82. Since the calculated t value is 10 which exceeds both critical values, the null hypothesis is rejected at both levels. Therefore, the claim that therapy has improved behaviour, is accepted.

7. List the assumptions of analysis of variance.

a) The variables described are expressed in interval or ratio scales.
b) The samples have equal or nearly equal variances.
c) Observations are independent

8. Thirty subjects were randomly assigned to three groups. Each group performed the same task under three experimental treatments. The scores of the subjects on the task are as under:

| Group I | Group II | Group III | 

| 26, 27, 18, 22, 23, 19, 27, 26, 24, 26 | 18, 22, 18, 23, 19, 24, 20, 21, 19, 25 | 20, 21, 25, 27, 28, 26, 25, 24, 20, 24 |

Test if there is a significant difference between the means of the three groups at the .05 level.

The calculated F value = 1.16. Since the table value is greater than the calculated value, the null hypothesis is accepted. The difference between the means of three groups is not significant.

9. List the advantages of converting Pearson's r into Fisher's Z . 

Following are the advantages of converting Pearson's into Fisher's Z: 

a) The sampling distribution ofZ coefficient is normal regardless ofthe size of sample N. 

b) The standard error of Z depends only upon the size of sample N. 

10. Find the .95 confidence interval for population Pearson's correlation coefficient of 0.78 obtained from a random sample of 84 cases. 

0.68 and 0 85

11. Test the significance of r for the following: r = 0.40 N = 25 

tr = 2.08, which is not significant at .05 level. Hence r = 0.40 is not significant 

12. Explain the meaning of statistical inference and its use in research?

Statistical inference refers to the process of drawing conclusions about a population based on data from a sample. It uses sample data to make generalisations or inferences about the population.

13. What are the assumptions of parametric tests?

• Variables described are expressed in interval or ratio scales.
• Samples have equal or nearly equal variances.
• Observations are independent.

14. How do you test the significance of difference between the means of two independent and dependent samples?For independent samples, a t-test is used to compare the means of two separate groups.

For dependent samples, a paired t-test is used to compare means of the same group at different time points.

15. What do you understand by analysis of variance? In what situations do you use this technique?

Analysis of variance (ANOVA) is a statistical technique used to test the differences between the means of three or more groups. It is used when comparing multiple groups to see if any of the group means are significantly different from each other.

16. Explain the use of Pearson's coefficient of correlation. What are the techniques for testing the significance of difference between two correlation coefficients?

Pearson's coefficient of correlation measures the strength and direction of a linear relationship between two continuous variables. The significance of the difference between two Pearson's correlation coefficients can be tested using Fisher's Z transformation.

17. Describe the characteristics of Central Limit Theorem. 

Central Limit Theorem Characteristics

The Central Limit Theorem describes the characteristics of sample means when a large number of equal-sized samples are selected at random from an infinite population. According to this theorem:

• The means of the samples will be normally distributed.
• The average value of the sample means will be the same as the mean of the population.
• The distribution of sample means will have its own standard deviation, which is known as the standard error of the mean.

18. Define the standard error of mean. 

Standard Error of the Mean Definition

The standard error of the mean (SEM), denoted as SEM or σM, is the standard deviation of the sampling distribution of the means. It indicates how far the sample means are expected to deviate from the population mean. It is calculated using the formula:

σM = σ / √N

where:

• σ is the standard deviation of the population.
• N is the size of the sample.

19. Define the confidence intervals and levels of confidence. 

Confidence Intervals and Levels of Confidence Definitions

Confidence intervals (or fiduciary limits) are ranges that estimate the interval within which the true population parameter is likely to fall. They help in adopting specific levels of confidence.

• Levels of confidence are the probability that the true population parameter lies within the confidence interval.
• Two common levels of confidence are:
  • 95% level (or .05 level): There is a 95% probability that the population mean falls within the interval M ± 1.96 σM, and a 5% chance it falls outside these limits.
  • 99% level (or .01 level): There is a 99% probability that the population mean falls within the interval M ± 2.58 σM, and a 1% chance it falls outside these limits.

20. Given a sample with mean = 34.85, SD=6.35 and N=100. Compute the .95 and .99 confidence intervals for the true mean. 

Calculating Confidence Intervals

For a sample with a mean of 34.85, SD of 6.35, and N of 100:

First, calculate the standard error of the mean (σM):

σM = 6.35 / √100 = 6.35 / 10 = 0.635

• .95 Confidence Interval:
  • 34.85 ± (1.96 * 0.635) = 34.85 ± 1.2446 = 33.61 to 36.09 (approximately)
• .99 Confidence Interval:
  • 34.85 ± (2.58 * 0.635) = 34.85 ± 1.6383 = 33.21 to 36.49 (approximately)

21. The mean of 12 independent observations of a test is 80 and SD is 14. Compute the .99 confidence intervals for the true mean. 

For a sample with a mean of 80, SD of 14 and N of 12:

• First, calculate the standard error of the mean. However, because this is a small sample, we will use the sample standard deviation (s) to calculate the standard error of the mean: sM = s / √N sM = 14 / √12 = 14 / 3.46 = 4.05

• Next we need the t-value for a 99% confidence interval, with 11 (N-1) degrees of freedom, which is 3.11.

• .99 Confidence Interval

  • 80 ± (3.11 * 4.05) = 80 ± 12.6 = 67.4 to 92.6 (approximately)

22. State the assumptions of Analysis of Variance. 

Analysis of Variance Assumptions

The basic assumptions underlying the analysis of variance are:

• The population distribution should be normal, though this is not especially critical.
• All groups should be randomly chosen from the relevant sub-populations. Randomness is a key assumption.
• Homogeneity of variance.

23. The Pearson's coefficient of correlation between the height and weight of 100 tenth grade boys is 0.85 and that of 75 tenth grade girls is 0.730. Is the difference between the correlations significant at .01 level?

Significance of Difference Between Correlations

To determine if the difference between Pearson's correlation coefficients of 0.85 (boys) and 0.73 (girls) is significant at the .01 level, we need to use the Fisher's Z transformation:

1. Convert the correlation coefficients into Fisher's Z coefficients. (Note: The source text does not include a conversion table so this cannot be computed.)
2. Calculate the standard error of the difference between Z coefficients.
3. Compute the critical ratio (CR) or Z value.
4. Compare the obtained Z value to the critical value at the .01 level of significance, which is 2.58.

The significance of the difference between the two r values, 0.80 and 0.85 from samples of M.Sc. students is tested in the source, giving a CR value of 1.65 which was not significant at .05 .