Business Statistics and Research Methods Questions and Answers

Answer: B

The F test cannot be used to test a hypothesis about a single population variance, which is done with the chi-square test. The F test can compare more than two population means in analysis of variance, it can test the equality of two population variances as the ratio of the sample variances, and it is the test of a randomized block design. So only option B describes a situation where F is not used. A situation with a single variance and a known standard value is where the chi-square distribution applies, so the F test does not apply there.

Answer: A

The paired 't' test is applied to test the effectiveness of special coaching, because the marks of the same students are taken before and after the coaching, and the observations are paired. A chi-square test is for frequencies, a Z test needs large samples or a known variance, and the F test compares variances or several means. The paired t test is made for before-and-after measurements on the same group. A before and after change in marks for the same students is the pairing, and the null hypothesis is that the mean difference is zero.

Answer: B

In the linear equation \(y = a + bx\), 'a' is the Y-intercept, the value of y when x is zero, where the line cuts the Y axis. The slope is b, which shows the change in y for a unit change in x. The dependent variable is y and the independent variable is x, so 'a' is neither of those.

Answer: D

The standard error of the sampling distribution of the mean is the standard deviation divided by the square root of the sample size.

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{30}{\sqrt{36}} \]

\[ SE = \frac{30}{6} = 5.0 \]

The mean of 220 is not needed for this calculation. The values 7.3, 6.2 and 6.0 correspond to using a different sample size or dividing by a different divisor, so option D, 5.0, is correct.

Answer: D

The number of ways of forming the committee with exactly two women, which means two of the four women and four of the seven men, is divided by the total number of committees of six from eleven people.

\[ P = \frac{\binom{4}{2}\binom{7}{4}}{\binom{11}{6}} \]

\[ P = \frac{6 \times 35}{462} = \frac{210}{462} \approx 0.4545 \]

The value 0.4545 is not the 0.456 listed in option A, and 0.803 and 0.962 are far from it, so None of the above, option D, is right.

Answer: A

The F value is the mean square between columns divided by the mean square of the remaining variation, here the rows, with degrees of freedom 3 and 16.

\[ MS_{columns} = \frac{100}{3} = 33.33 \]

\[ MS_{rows} = \frac{24}{16} = 1.5 \]

\[ F = \frac{33.33}{1.5} = 22.2 \]

Option C, 33.3, is only the numerator mean square, and the other options do not arise from the table.

Answer: B

The power of a test is 1 minus beta, the probability of rejecting a false null hypothesis, so with beta = 0.10 the power is 0.90.

The value 0.95 is 1 minus alpha, the confidence level, not the power. The value 0.80 and 0.15 do not follow from the given alpha and beta, since 0.15 is just their sum.

Answer: A

The chi-square value is 2.67.

Chi-square is the sum of (observed - expected) squared over expected for each event.

\[ \chi^{2} = \frac{(52-60)^{2}}{60} + \frac{(48-40)^{2}}{40} \]

\[ = \frac{64}{60} + \frac{64}{40} = 1.067 + 1.600 = 2.67 \]

So option A is correct; 2.77, 3.33 and 4.33 do not result from the data. The result is also a check on the arithmetic: the two terms are 1.067 and 1.600, and the nearest of the four options is 2.67, so the other values cannot be the answer.

Answer: B

A time series is a set of data collected at regular intervals.

The observations are recorded at equal, fixed gaps of time such as days, months or years, so that trend, seasonal and cyclical movements can be measured. Random or merely convenient intervals do not give a consistent series, and so options A, C and D are not correct. Without equal spacing the movements of the series cannot be separated into trend, seasonal variation, cyclical swings and irregular changes, which is the main use of the series in forecasting, and so the interval must be regular and not random or merely convenient.

Answer: C

The regression coefficient of X on Y is 0.80.

The correlation coefficient is the geometric mean of the two regression coefficients, so the square of r equals the product of b_yx and b_xy.

\[ r^{2} = b_{yx} \times b_{xy} \]

\[ 0.64 = 0.8 \times b_{xy} \]

\[ b_{xy} = \frac{0.64}{0.8} = 0.80 \]

The values 0.32, 0.64 and 0.82 do not satisfy this relation, so C is correct. The shortcut to remember is that the regression coefficients have the same sign as the correlation, and their product equals r squared, so when one of them and r are known the other follows by one division.