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Unit 6 - Week 4
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How to access the portal Assignment 00 Week 1 Week 2 Week 3 Week 4
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Lecture 16:
MLE Lecture 17:
STATISTICAL INFERENCE Lecture 18:
HYPOTHESIS TESTING Lecture 19:
HYPOTHESIS TESTING Lecture 20:
HYPOTHESIS TESTING Quiz : Assignment 4 WEEK 4 - FEEDBACK - Data Analysis and Decision Making - I Assignment 4 Solution
Due on 2018-09-05, 23:59 IST.
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Assignment 4
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Upper and lower boundaries of confidence interval are termed as:
error limits estimate limits confidence limits marginal limits No, the answer is incorrect.
Score: 0
Accepted Answers:
confidence limits
A random sample of 150 Students from MSc Statistics course in a central university are
given a test to judge their proficiency in advanced statistics, and 60 of these people had levels over the
“required threshold" marks of 200. Construct a 95% confidence interval for the population proportion of people in the university with marks over 200.
Lower limit = 0.3216, and upper limit = 0.4784 Lower limit = 0.333, and upper limit = 0.4784 Lower limit = 0.3456, and upper limit = 0.4784 Lower limit = 0.3166, and upper limit = 0.4784 No, the answer is incorrect.
Score: 0
Accepted Answers:
Lower limit = 0.3216, and upper limit = 0.4784
Confidence intervals are a function of which of the following three things?
The population, the sample, and the standard deviation The sample, the variable of interest, and the degrees of freedom The data in the sample, the confidence level, and the sample size
The sampling distribution, the confidence level, and the degrees of freedom No, the answer is incorrect.
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The data in the sample, the confidence level, and the sample size
Marketing manager of Coke is assigned a task to find whether more people prefer Coke to Pepsi. Assume that roughly half of the population prefers Coke and half prefers Pepsi. What size of the sample the manager needs to estimate, with 95% confidence, the proportion of people who prefer Coke within 3% of the actual value?
1068 1168 1086 1608
No, the answer is incorrect.
Score: 0
Accepted Answers:
1068
Which of the following statements is FALSE?
when sample data are used for estimating a population mean, sampling error is not be present because the observed sample statistic is same as the actual value of the population parameter
Inferential statistics is the process of using sample statistics to estimate population parameters
a point estimate is a single value estimate of the value of a population parameter
an interval estimate is an estimate of the range of possible values for a population parameter No, the answer is incorrect.
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when sample data are used for estimating a population mean, sampling error is not be present because the observed sample statistic is same as the actual value of the population parameter
An marketer of new smartphone in market is believed to exaggerate claims about a
company's product, (higher performance, longer average battery life). An independent agency wants confirm whether this advertiser's claims are exaggerated. you assume that all the required data available. The correct hypothesis test will be:
Two-tailed test Right-hand tailed test Left-hand tailed test None of the above No, the answer is incorrect.
Score: 0
Accepted Answers:
Left-hand tailed test
An interval's confidence level ___________, provides an estimate of the likelihood that it contains the ___________ of interest.
(1-a)*100%; population parameter (1-a)*100%; sample statistic (1-a/2)*100%; population parameter
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(1-a/2)*100%; sample statistic No, the answer is incorrect.
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(1-a)*100%; population parameter
If the mean of sampling distribution is equal to population mean then the sample statistic is classified as
unbiased estimator biased estimator interval estimator error estimator No, the answer is incorrect.
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unbiased estimator
The level of confidence is denoted by:
α 1-α β 1-β
No, the answer is incorrect.
Score: 0
Accepted Answers:
1-α
Increasing (1 – α) ________ the width of a confidence interval:
Decreases Increases
Brings no change in Flattens
No, the answer is incorrect.
Score: 0
Accepted Answers:
Increases
In a t-distribution for two independent samples n1 = n2 = n. Calculate the degrees of freedom.
2n-1 2n+1 2n-2 2n+2
No, the answer is incorrect.
Score: 0
Accepted Answers:
2n-2
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A 90% confidence interval for the mean of a population is such that:
It contains 90% of the values in the population
There is a 90% chance that it contains all the values in the population.
There is a 90% chance that it contains the mean of the population
There is a 90% chance that it contains the standard deviation of the population No, the answer is incorrect.
Score: 0
Accepted Answers:
There is a 90% chance that it contains the mean of the population
A random sample of 30 engines produced in a factory in March 2018 was selected as part of a study on their efficiency, and the horsepower was recorded for each engine in the sample. The average engine power was found to be 375hp. But in a very large study it was found that the standard deviation of the engine power was 81hp. If we assume that the standard deviation is unchanged and that the efficiency is normally distributed, what is the 99% confidence interval for the mean power in March 2018.
375 ± 2.756×(81/√30) 375 ± 2.576×(81/√30) 375 ± 2.657×(81/√30) 375 ± 2.575×(9/√30) No, the answer is incorrect.
Score: 0
Accepted Answers:
375 ± 2.575×(9/√30)
An experiment was conducted to understand whether or not an observer’s brain signals
respond differently between print advertisements in BLUE and RED on an average. Brain activity was measured while 17 patients were exposed to BLUE and RED advertisements. The differences (BLUE- RED) in the equipment as obtained were d1, d2, … d17 and ∑di=−3.50 and ∑di*di=19.13 (where ∑ is the summation operator). Provide a 90% confidence interval for the average difference (BLUE- RED).
(-0.21, 0.21) (-0.66, 0.25) (-0.76, 0.35) (-0.63, 0.43)
No, the answer is incorrect.
Score: 0
Accepted Answers:
(-0.66, 0.25)
A random sample of 100 children in the age group 1 month - 12 months in Kanpur shows
that only 60 kids had been vaccinated. Develop a 95% confidence interval for the proportion of children vaccinated in that age-group in Kanpur.
0.6 ± 1.96(√((0.6×0.4)/100)) 0.6 ± 1.65(√((0.6×0.4)/100)) 0.6 ± 1.95(√((0.6×0.4)/100)) 0.6 ± 1.32(√((0.6×0.4)/100)) No, the answer is incorrect.
Score: 0
Accepted Answers:
0.6 ± 1.96(√((0.6×0.4)/100))
