ECON634 Econometrics and Business Statistics For Population Parameter
Questions:
2. Assume that 50% of students scored above 493 in the numeracy test. Calculate the probability that, in a randomly selected sample of 200 Year 5 children, more than 115 of them scored over 493 in the numeracy test. Justify any assumption
you make. Shade in the diagram to represent the probability you have calculated.
3. It is reported that the average numeracy score achieved by Year 5 classes in the Naplan tests of 2016 is 493. From the sample of 59 schools in 2015, the average score was found to be 506.63 with a standard deviation of 36.76. Test the hypothesis that the average numeracy scores of Year 5 classes in 2015 in the three major capital cities differs from the average numeracy score in 2016. Test your claim at the 5% significance level.
5. Calculate a 95% confidence interval to estimate the average numeracy score of all grade 5 classes in the major capital cities in 2015. Interpret this interval.
Answer:
The distribution of numeracy scores was assumed to be normally distributed using central limit theorem, as the sample size was greater or equal to 30.
Here, and
The required probability was calculated.
The distribution of numeracy score was assumed to be normally distributed using central limit theorem, as the sample size was greater than 30.
According to the problem, population parameter ratio for numeracy score was
Now, sample size
Required probability was
Null hypothesis: H0:
Alternate Hypothesis: HA: where was the average numeracy score achieved by Year 5 classes.
Level of significance was considered at 5%, and the test was a two tail test.
Sample statistic values were
Due to absence of population standard deviation, t-test was the appropriate choice.
The t-statistic = with 58 degrees of freedom.
The p-value was < 0.05, at 5% level.
The 95% confidence interval for population mean was estimated as, and the population average was outside the interval.
Hence, the null hypothesis was rejected at 5% level of significance, concluding that average numeracy scores in the three major capital cities were significantly different from the average numeracy score of 493.
The 95% confidence interval for average numeracy score of all grade 5 classes in the major capital cities in 2015 was estimated as, and the population average was found to be outside the interval. Hence, the null hypothesis was correctly rejected at 5% level of significance.
Null hypothesis: H0:
Alternate Hypothesis: HA:
Level of significance was considered at 5%, and the test was a two tail test.
Sample statistic values were
Z-test was the appropriate choice.
The Z-statistic =
The p-value was < 0.05, at 5% level.
The 95% confidence interval for population mean was estimated as, and the population parameter was outside the interval.
Hence, the null hypothesis was rejected at 5% level of significance, concluding that proportion of numeracy scores in the three major capital cities were significantly different from the population proportion of numeracy score.
The 95% confidence interval for population mean was estimated as, and the population parameter was outside the interval. Hence, the null hypothesis was correctly rejected at 5% level of significance,
Sample statistic values were
The 95% confidence interval for population mean was estimated as, and the population parameter was still outside the interval. Hence, the null hypothesis would be still rejected at 5% level of significance,
Null hypothesis: H0:
Alternate Hypothesis: HA: where and were the variances of Writing scores of Year 3 children from Sydney schools and Melbourne schools in 2015.
Level of significance was considered at 5%, and the test was a two tail test.
Sample statistic values were
The F-test was the appropriate choice. The F-test was performed in Excel to test the hypothesis at 5% level.
The F-calculated was 0.92, and p-value = 0.428 > 0.05, at 5% level.
Hence, the null hypothesis failed to get rejected at 5% level of significance, concluding that there was no statistically significant difference in variances of writing scores of Year 3 children from Sydney schools and Melbourne schools in 2015.
Null hypothesis: H0:
Alternate Hypothesis: HA:
Level of significance was considered at 5%, and the test was a two tail test.
Sample statistic values were
Due to absence of population standard deviations, t-test with equal variances (from 2.b) was the appropriate choice.
The p-value was > 0.05, at 5% level.
Hence, the null hypothesis failed to get rejected at 5% level of significance, concluding that there was no statistically significant difference in the average writing scores of Year 3 children from Melbourne schools and Sydney schools in 2015.
The p-value was > 0.05, for two tail test at 5% level.
Hence, the null hypothesis failed to get rejected at 5% level of significance, concluding that there was no statistically significant difference in the average writing scores of Year 3 children from Melbourne schools and Sydney schools in 2015.
The histogram was symmetric in nature, and the shape almost resembled to that of the normal distribution (bell shaped curve).
Null hypothesis: H0:
Alternate Hypothesis: HA: where and were average Reading and Writing scores.
Level of significance was considered at 5%, and the test was a right tail test.
Sample statistic values were
Due to absence of population standard deviations, t-test with unequal variances was the appropriate choice.
The t-statistic = 2.197 with 35 degrees of freedom.
The p-value was < 0.05, at 5% level.
Hence, the null hypothesis was rejected at 5% level of significance, concluding that the average Reading scores were statistically significantly more than the average writing scores of Year 3 classes (Fang, 2017).
The 95% confidence interval for average numeracy score of mean difference in average Reading scores and average Writing scores of Year 3 classes was estimated as, and the difference in average Reading scores and average Writing scores was found to be outside the interval. Hence, the null hypothesis was correctly rejected at 5% level of significance.
Excel was used to estimate the regression line of Grammar on Numeracy scores. The estimated regression line was, where was the residual of the estimated equation
The estimated regression line was, where was the residual of the estimated equation
The standard error for intercept was SE = 69.813, and that of the Numeracy scores was SE = 0.133.
The intercept (t = -0.067, p = 0.948) was statistically insignificant, as negative Grammar score due to zero in Numeracy scores does not make any statistical sense.
Numeracy was statistically significant (t = 7.82, p < 0.05) predictor of Grammar scores at 5% level of significance.
The sample size was 19, and the coefficient of determination (R-square) = 0.782. Hence, Numeracy was able to explain 78.2% variation in Grammar scores.
The slope coefficient was 1.038, which signified a significant strong positive linear relation between the two variables. The angle of the regression line was noted to be greater than 45 degrees, as the slope was greater than 1.
Null hypothesis: H0: (No linear relation exists)
Alternate Hypothesis: HA:
Level of significance was considered at 5%, and the test was a two tail test.
Sample statistic values were
Due to absence of population standard deviation, t-test was the appropriate choice.
The t-statistic = with 18 degrees of freedom.
The p-value was < 0.05, at 5% level.
The 95% confidence interval for population mean was estimated as, and the population parameter was outside the interval.
Hence, the null hypothesis was rejected at 5% level of significance, concluding that a statistically significant linear relation exists between Grammar scores and Numeracy scores of Year 5 classes in Sydney in 2015 (Chatterjee, & Hadi, 2015).
References
Chatterjee, S., & Hadi, A. S. (2015). Regression analysis by example. John Wiley & Sons.
Fang, K. W. (2017). Symmetric Multivariate and Related Distributions: 0. Chapman and Hall/CRC.
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