A survey was conducted to study the preference for online learning among students from urban and rural areas. Out of 300 urban students, 210 preferred online learning, while out of 250 rural students, 150 preferred online learning. Construct a 95% confidence interval for the difference between the population proportions of students preferring online learning in urban and rural areas.
Let ( p_1 ) be the proportion of urban students preferring online learning, and ( p_2 ) be the proportion of rural students preferring online learning. From the data: [ n_1 = 300, \quad x_1 = 210 \implies \hat{p}_1 = \frac{210}{300} ____ _________ __________ _____ ___ _________ __________ _______ ____.
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A researcher wants to compare the average test scores of students taught using Method A and Method B. A random sample of students was selected from each group, and their scores are given below:
| Student | Method A | Method B |
|---|---|---|
| 1 | 68 | 72 |
| 2 | 74 | 75 |
| 3 | 71 | 78 |
| 4 | 69 | 70 |
| 5 | 73 | 76 |
| 6 | 70 | 74 |
| 7 | 72 | 77 |
| 8 | 75 | 79 |
Let X1, X2, ..., Xbe a random sample from the Exponential distribution with pdf f(x; 0) = 0e-ex, x > 0, 0 > 0
(i) Obtain the maximum likelihood estimator (MLE) of 0.
(ii) Hence, find the maximum likelihood estimate of for the observed sample: 2.1, 1.6, 3.4, 0.9, 2.5
State whether the following statements are True or False. Give reasons in support of your answer:
(i) The mean square error is calculated when the estimator is unbiased.
(ii) If the sample size tends to infinity, the variance of a consistent estimator must tend to zero.
(iii) In the method of moments, population moments are equated to corresponding sample moments.
(iv) A 95% confidence interval is smaller than a 99% confidence interval.
(v) The chi-square distribution is symmetric for all degrees of freedom.
Describe the following:
(i) t-distribution
(ii) General procedure of Testing of hypothesis
A school caretaker looks after 6 children. The ages (in years) of the children are given below:
| Child | Age (in years) |
|---|---|
| Meena | 5 |
| Aarav | 4 |
| Nisha | 6 |
| Kabir | 4 |
| Pooja | 5 |
Let be a random sample from a population with mean
and variance
. Consider the estimator
$
(i) Find the bias of the estimator .
(ii) Find the variance of .
(iii) Hence, calculate the Mean Square Error (MSE) of .
A survey was conducted to study the preference for online learning among students from urban and rural areas. Out of 300 urban students, 210 preferred online learning, while out of 250 rural students, 150 preferred online learning. Construct a 95% confidence interval for the difference between the population proportions of students preferring online learning in urban and rural areas.