In each of the multiple choice questions ONE of the answers is CORRECT, except where stated otherwise (and then ONE is WRONG).

Q1. When a 95% confidence interval is constructed for a parameter θ

the interval contains q with probability 0.95
θ lies in the interval with probability 0.95
95% of the intervals constructed contain the estimate       of θ found from the data
the interval covers 95% of the possible values for θ
the calculation uses the middle 95% of the data.

Q2. When a 95% confidence interval is calculated for the mean µ of a normal distribution, using the usual formula and a sample of n observations

the interval is symmetrical about µ
the interval is of width 2σ²/n
the calculation uses the middle 95% of the data
the calculation uses t with n degrees of freedom
the interval is symmetrical about the mean of the given data

Q3. A 95% confidence interval for a proportion p based on 250 observations is (0.35, 0.54). This shows that

p cannot be less than 0.35
p cannot be more than 0.54
with 95% probability, the true value of p lies between        0.35 and 0.54
p is significantly greater than 0.4
p is significantly different from 0.5

Q4. A 95% confidence interval for a variance in a normal distribution is (15.65, 95.85). The corresponding 95% confidence interval for the standard deviation is

(1.13, 14.02)
(3.96, 9.79)
(6.04, 8.84)
(7.56, 10.71)
(3.00, 6.28)

Q5. When estimating both the mean and the variance of a normal distribution using a random sample of data containing n observations, the maximum likelihood estimator of the variance is

the same as the method of moments estimator
unbiased
the same as it would be if the mean were known
(n – 1)/n times the true variance
equal to the true variance

 

Q6. The χ² random variable

takes all values from - ∞ to ∞
has a skew distribution
can be used in tests of medians
only tests data from discrete distributions
gives only approximate tests.

Q7.In a χ² test for a contingency table having 3 rows and 4 columns

The binomial distribution may not be a suitable model for these data because:

there are 11 degrees of freedom
expected frequencies should be expressed in whole       numbers, the same as the observed
a significant result shows there is no relation between       row and column classifications
there are 6 degrees of freedom
the original data must be qualitative.

Q8. The sum of independent geometric variables follows

a negative binomial
a binomial
a geometric
a Poisson
no simple distribution

Q9. The correlation coefficient between x and y shows

whether y depends on x
whether x causes y
whether there is any relation between x and y
whether there is a linear relation between x and y
whether positive values of x go with negative values of y

Q10. A value of –0.55 for a correlation coefficient calculated from 42 pairs of data suggests that

there is no relation between x and y
x and y decrease together
x increases as y decreases
the relation between x and y is curved
there has been an error in calculation

Q11. A significant value of a correlation coefficient calculated from a sample of data (x, y)
implies that

x causes y
y causes x
there is a third variable which explains both of x, y
x and y have a curved relationship
x and y have a relationship with a strong linear component.

Q12.Rank correlation should NOT be used when

the pairs of data are measured as continuous variables
there are no outliers in the data
the pairs of data are jointly normally distributed
data are measures as integer (discrete uniform) variables
sample size is small.

Q13. X has mean µ_X and variance σ_X²; Y has mean µ_y and variance σ_y². The correlation coefficient between them is ρ. The covariance of (aX + Y) and (X + bY) is

a²σ_X² + 2ρabσ_Xσ_y + b2σ_y2
σ_X2 + ρσ_Xσ_y + sY2
a2σ_X2 + ρabσ_Xσ_y + b2σ_y2
aσ_X2 + (1 + ab)ρσ_Xσ_y + bσ_y2
aσ_X2 + ρabσ_Xσ_y + bσ_y2.

Q14. If R follows a Poisson distribution with mean λ

the mean and standard deviation of R are equal
the possible values of R are 0, 1, 2, …, n
P(R = 0) = λP(R – 1)
P(R = 0) = e-^λ
R describes a random event which occurs at a steadily decreasing rate over time.

Q15.In a Poisson distribution with mean 2, the probability P(R ≥ 3) is

0.677
0.271
0.323
0.541
0.857

Q16.The monthly number R of breakdowns in a large computer system follows a Poisson distribution with mean 3. The probability of observing more than 4 but not more than 7 breakdowns during one month is

Poisson with mean 3.84
Normal with mean 4.8 and variance 2.4
Poisson with mean 4.0
binomial with p = 2/3
not a simple distribution.

Q17.The weekly average number of a particular type of accident treated at a hospital has been 4.4, based on data for a large number of weeks. A week could be regarded as unusual if there were :

0
1
2
4
7

of these accidents.

Q18. A set of 100 observations is obtained on r which is thought to follow a Poisson distribution. The mean and variance, or standard deviation, of the data are calculated. The Poisson hypothesis is a reasonable one if

mean = 2.05, standard deviation = 2.35
mean = 2.23, variance = 9.54
mean = 1.48, variance = 1.62
mean = 7.35, standard deviation = 4.39
mean = 4.10, variance = 16.81.

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