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} = 0.7 ] [ n_2 = 250, \quad x_2 = 150 __________ ____ ________ ___ ____ ______ ______ _________ ______ ____ ____ __________.
__________ _________ ______ _______ _________ ________ ________ __________.
_______ ___ ____ ________ ___ _________ _______ _________ __________ ______ _____ __________.
_______ ________ ____ ____ ________ __________ ____ ______ _____ ___.
_________ ____ ______ _____ ______ __________ _______ ________ __________ ________ ____.
____ ________ ________ __________ _____ ____ _________ ___ ______ _____ ___ _________.
________ ___ _________ _______ _________ _________ ________ __________ ____ _____.
_______ ___ __________ ____ ________ _______ ______ ___ ___ __________ ____ ______.
___ ___ ________ ________ _________.
____ _____ _______ ______ _____ _____.
_______ _________ _________ _________ _______.
___ ____ ______ __________ _____ _____ ____ _________.
__________ _______ _________ __________ ___ ______.
________ ________ _________ __________ _________ _________ ___ __________ __________ ________ _________ ________.
____ _______ _____ _____ ______ ______.
__________ __________ _____ ______ ________ ________.
________ ____ ________ ______ ____ __________ _________ _____ ___.
__________ _______ _________ ___ ________ __________.
Get Full Answer on WhatsApp
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 |
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.
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.
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 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 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
Describe the following:
(i) t-distribution
(ii) General procedure of Testing of hypothesis