Business Statistics and Research Methods Questions and Answers

Answer: C

P(A' ∩ B) = 0.5.

The event B splits into the part that is also in A and the part that is not:

\[ \begin{aligned} P(B) &= P(A \cap B) + P(A' \cap B) \\ P(A' \cap B) &= 0.8 - 0.3 = 0.5 \end{aligned} \]

P(A) is not needed for this result. 0.1 is P(B) minus P(A), and −0.1 would be impossible as a probability. 0.7 is simply P(A) repeated. (Note that P(A) + P(B) - P(A ∩ B) = 1.2 is greater than 1, so the printed figures are not fully consistent, but the formula above uses only P(B) and P(A ∩ B).)

Answer: C

The parameters are n = 16, p = 0.5.

For a binomial distribution the mean is np and the variance is npq, with q = 1 - p:

\[ \begin{aligned} np &= 8,\quad npq = 4 \\ q &= 4/8 = 0.5 \\ p &= 0.5,\quad n = 8/0.5 = 16 \end{aligned} \]

Check: 16 \(\times\) 0.5 = 8 and 16 \(\times\) 0.5 \(\times\) 0.5 = 4. The pair n = 8, p = 0.5 gives a mean of only 4, and the pairs with p = 0.4 give a mean of 1.6 or 6.4, so none of them matches.

Answer: C

The area between −1.96 and +1.96 is 95%.

From the standard normal table, the area from 0 to 1.96 is 0.4750. By symmetry the area from −1.96 to 0 is the same, so the total is 0.95, leaving 2.5 percent in each tail. This is why 1.96 is the critical value of z for a two-tailed test at the 5 percent level and for a 95 percent confidence interval.

5% is the area outside the two lines, not between them. About 90% lies between −1.645 and +1.645, and 50% lies on either side of the mean, so those options belong to other z values.

Answer: B

The test is the t test.

To compare the means of two groups when the population standard deviations are unknown, they are estimated from the samples, and the difference of means divided by its estimated standard error follows Student's t distribution (with n₁ + n₂ - 2 degrees of freedom when the variances are pooled). This is the two-sample t test, used especially for small samples.

The z test needs known population standard deviations, or large samples. The F test compares variances, not means, and chi-square tests goodness of fit or association between attributes.

Answer: C

ANOVA involves splitting of the variance as variation between the samples and also within sample.

R.A. Fisher's analysis of variance divides the total variation of all the observations into two parts: the variation between the sample means, which reflects differences between the groups, and the variation within each sample, which reflects chance. The ratio of the two mean squares is the F statistic used to test whether the group means are equal.

Computing the means and variances of each group, or setting out the ANOVA table, are steps in carrying out the analysis, but they do not describe what the technique essentially does.

Answer: B

It is called Applied research.

Applied (action-oriented) research aims to find a solution to an immediate, practical problem facing a society or an organisation, for example research into why a firm's sales are falling or how to reduce absenteeism.

Fundamental (basic or pure) research seeks generalisations and theory, knowledge for its own sake, without a specific problem to solve. Analytical research uses facts already available and analyses them critically, and conceptual research deals with abstract ideas and theories. Neither is defined by an immediate practical problem.

Answer: B

The correct order is (ii), (iii), (i), (v), (iv).

The standard steps of report writing (as set out by C.R. Kothari) are:

1. Logical analysis of the subject matter (ii).
2. Preparation of the final outline (iii).
3. Preparation of the rough draft (i).
4. Rewriting and polishing of the rough draft (v).
5. Preparation of the final bibliography (iv), followed by writing the final draft.

The other orders put a draft before the analysis and outline that it depends on, or polish a draft before it exists, so they cannot be right.

Answer: B

The reasons are (i), (iii), (iv), (v).

The arithmetic mean is popular because it uses every observation (i), lends itself to further algebraic treatment, for instance combining group means (iii), is rigidly defined by a formula so everyone gets the same value (iv), and fluctuates less from sample to sample than most other averages (v).

Statement (ii) is false. The mean is most affected by extreme observations, because every value enters the sum; the median and mode are the averages that resist outliers. Every option containing (ii) is therefore wrong, and (i), (v) alone leaves out two genuine merits.

Answer: C

The standard deviation is 0.

The mean of 7, 7, 7, 7, 7 is 7, so every deviation from the mean is zero:

\[ \begin{aligned} \sigma &= \sqrt{\frac{\sum (x - \bar{x})^{2}}{n}} \\ &= \sqrt{\frac{0}{5}} = 0 \end{aligned} \]

Standard deviation measures spread, and a set of identical values has no spread at all. 7 is the mean, not the dispersion, and 49 and the square root of 7 come from squaring or rooting the value itself, which is not how dispersion is calculated.

Answer: A

\(P(A \cup B) = \) 0.94.

For independent events, \(P(A \cap B) = P(A)P(B)\), and the addition rule gives:

\[ \begin{aligned} P(A \cap B) &= 0.7 \times 0.8 = 0.56 \\ P(A \cup B) &= 0.7 + 0.8 - 0.56 = 0.94 \end{aligned} \]

A check: the probability that neither occurs is \(0.3 \times 0.2 = 0.06\), and \(1 - 0.06 = 0.94\).

0.56 is the probability of both A and B, not of either. 0.75 is below P(B) = 0.8, which a union can never be, and 0.80 is just P(B), which would mean A adds nothing outside B; for independent events that is impossible here, since A occurs without B with probability 0.7 × 0.2 = 0.14.