Business Statistics and Research Methods Questions and Answers

Answer: B

The answer is C.V. (X) > C.V. (Y).

The coefficient of variation is the standard deviation as a percentage of the mean:

\[ \begin{aligned} CV_{X} &= \frac{10}{100} \times 100 = 10\% \\ CV_{Y} &= \frac{15}{200} \times 100 = 7.5\% \end{aligned} \]

Although Y has the larger standard deviation, X varies more relative to its mean, so X is the less consistent series. The ratio of the two is 10/7.5, about 1.33, not 1.5, so the last option is also wrong.

Answer: C

P(A) = 2/3.

Mutually exclusive and exhaustive events cannot occur together and together make up the whole sample space, so their probabilities add to 1.

\[ \begin{aligned} P(A) + P(B) &= 1 \\ 2P(B) + P(B) &= 1 \Rightarrow P(B) = \tfrac{1}{3} \\ P(A) &= \tfrac{2}{3} \end{aligned} \]

1/3 is P(B), not P(A). 0 and 1 would make one of the events impossible, which contradicts P(A) = 2P(B) with both summing to 1.

Answer: D

The probability is \((0.8)^{5}\).

Treating the five jobs as independent, each completed in time with probability 0.8, the probability that all five succeed is the product of the five probabilities (the binomial probability of 5 successes in 5 trials):

\[ \begin{aligned} P &= 0.8 \times 0.8 \times 0.8 \times 0.8 \times 0.8 \\ &= (0.8)^{5} \approx 0.328 \end{aligned} \]

5 × 0.8 = 4 is the expected number of jobs completed, and is not even a probability. 0.8 is the chance for one job only, and 1 would mean completion is certain.

Answer: A

In systematic sampling the samples drawn are not overlapping.

In systematic sampling the population is listed and every k-th unit is taken after a random start between 1 and k. Each possible random start gives a different sample, and these k possible samples divide the population between them: every unit belongs to exactly one of them, so no two samples share a unit.

The method is random because the start is chosen at random, and it usually gives a well spread, representative sample unless the list has a periodic pattern matching k. So the other three options do not describe it.

Answer: A

The critical region is the region of rejection for \(H_{0}\).

In a test of hypothesis, the sample space of the test statistic is divided into two parts. If the statistic falls in the critical region, the null hypothesis \(H_{0}\) is rejected; otherwise it is not rejected. The size of the critical region is the level of significance, the probability of rejecting \(H_{0}\) when it is true.

The test is framed around \(H_{0}\), so the critical region is not described as the rejection region of \(H_{1}\), and it is the opposite of an acceptance region.

Answer: B

The Mann-Whitney U test is used when the samples are from non-normal populations.

It is a non-parametric test that compares two independent samples using the ranks of the pooled observations, so it does not require the populations to be normal. It is the standard alternative to the two-sample t test when the normality assumption fails.

If the populations are normal, the t test (unknown variances) or z test (known variances) is used instead. Unequal sample sizes do not call for the U test; both the t test and the U test handle them.

Answer: C

The means are x = 1, y = 2.

Both regression lines pass through the point of means, so solve them together.

\[ \begin{aligned} x &= 5 - 2y \\ 2(5 - 2y) + 3y &= 8 \\ 10 - y &= 8 \Rightarrow y = 2 \\ x &= 5 - 4 = 1 \end{aligned} \]

Check: 1 + 2(2) = 5 and 2(1) + 3(2) = 8. The pair 2, 1 reverses the two means, and -3, 4 and 2, 4 do not satisfy both equations.

Answer: C

P(X > 20) = 0.5.

A normal distribution is symmetric about its mean, which is also its median. With mean 20, exactly half the probability lies above 20 and half below, whatever the variance:

\[ P(X \gt 20) = P(Z \gt 0) = 0.5 \]

The variance of 64 (standard deviation 8) only affects how spread the distribution is. 20/64 wrongly mixes the mean with the variance; 0 and 1 would mean no values or all values lie above the mean.

Answer: C

The order is e, c, b, d, a.

1. Decide the sampling method (e).
2. Determine the size of the sample (c).
3. Draw the sample (b).
4. Conduct the survey to collect data (d).
5. Test the hypothesis on the data collected (a).

Hypothesis testing needs data, so it cannot come first, which rules out the order starting with a. The survey cannot be conducted before a sample is drawn, which rules out the order starting with d, and a sample cannot be drawn before its method and size are fixed.

Answer: B

The combined mean salary is ₹ 19,000 per month.

The combined mean weights each group's mean by its size:

\[ \begin{aligned} \bar{x} &= \frac{100 \times 20000 + 50 \times 17000}{150} \\ &= \frac{28{,}50{,}000}{150} = 19{,}000 \end{aligned} \]

₹ 18,500 would be the simple average of the two means, which ignores that there are twice as many men; it is not even an option. ₹ 20,000 and ₹ 17,000 are the two group means.