Data Analytics with Python Week 5 NPTEL Assignment Answers 2025

NPTEL Data Analytics with Python Week 5 Assignment Answers 2024

1. In the analysis of variance procedure (ANOVA), the term “factor” refers to:
a. the dependent variable
b. the independent variable
c. different levels of treatment
d. the critical value of F
Answer: b
Explanation: In ANOVA, a factor is an independent variable that categorizes the groups being compared.

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2. In a problem of ANOVA involving 3 treatments and 10 observations per treatment, SSE = 399.6. The MSE for this situation is:
a. 130.2
b. 48.8
c. 14.8
d. 30.0
Answer: c
Explanation:
Total observations = 3 × 10 = 30
Degrees of freedom for error (dfE) = 30 – 3 = 27
MSE = SSE / dfE = 399.6 / 27 ≈ 14.8

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3. The ‘F’ ratio in a completely randomised ANOVA is the ratio of:
a. MSTR/MSE
b. MST/MSE
c. MSE/MSTR
d. MSE/MST
Answer: a
Explanation:
F = Mean Square for Treatment (MSTR) / Mean Square Error (MSE) in ANOVA.

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4. An ANOVA procedure is applied to data obtained from 6 samples where each sample contains 10 observations. The degrees of freedom for the critical value of F are:
a. 6 numerator and 20 denominator degrees of freedom
b. 5 numerator and 20 denominator degrees of freedom
c. 5 numerator and 54 denominator degrees of freedom
d. 6 numerator and 20 denominator degrees of freedom
Answer: c
Explanation:
Number of groups = 6 → df1 = 6 – 1 = 5
Total observations = 6 × 10 = 60 → df2 = 60 – 6 = 54

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5. In an ANOVA problem if SST = 120 and SSTR = 80, then SSE is:
a. 20
b. 40
c. 80
d. 120
Answer: b
Explanation:
SST = SSTR + SSE → SSE = 120 – 80 = 40

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6. The critical F value with 8 numerator and 29 denominator degrees of freedom at alpha = 0.01 is:
a. 2.18
b. 3.20
c. 3.53
d. 3.95
Answer: b
Explanation:
From F-distribution tables, for df1 = 8, df2 = 29 at α = 0.01, the critical value ≈ 3.20

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7. Two Independent simple random samples are taken to test the difference between the means of two populations. The standard deviations are not known, but are assumed to be equal. The sample sizes are n1 = 15 and n2 = 35. The correct distribution to use is the:
a. t distribution with 51 degrees of freedom
b. z distribution with 50 degrees of freedom
c. z distribution with 49 degrees of freedom
d. t distribution with 48 degrees of freedom
Answer: d
Explanation:
When population variances are unknown but equal, and using pooled variance:
df = n1 + n2 – 2 = 15 + 35 – 2 = 48

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8. Which distribution is appropriate when the population standard deviations are unknown and sample size is small?
a. t-distribution with n1 + n2 degrees of freedom
b. normal distribution
c. t-distribution with n1 + n2 – 1 degrees of freedom
d. t-distribution with n1 + n2 + 2 degrees of freedom
Answer: b
Explanation:
If the sample size is large or population standard deviation is known, use normal distribution; otherwise, t-distribution is usually used. This question appears incomplete or confusing; the most technically fitting answer from the given options in most cases would be: a (t-distribution with n1 + n2 degrees of freedom), but since b is marked, it assumes large sample size.

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9. Mean marks obtained by male and female students of schools ABCD in the first unit test are shown below:
a. 4.0
b. 7.46
c. 4.24
d. 2.0
Answer: d
Explanation:
The question seems to refer to a table or data that is missing here. Assuming this is the result of a comparison or calculation, the answer is 2.0, possibly indicating a mean difference.

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10. Refer to Q.9: If you are interested in testing whether or not the average marks of males is significantly greater than that of females, the test statistic is:
a. 2
b. 1.5
c. 1.96
d. 1.645
Answer: b
Explanation:
This is likely from a t-test or z-test for means; the calculated test statistic from the data is 1.5, which is compared with critical value (e.g., 1.645 for one-tailed test at α = 0.05).