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STA260

Winter 2022

Term Test 1: Solutions

Q1: Let X1 , . . . , X9 denote a random sample from the standard normal distribution. Find a such that:

P ╱ ≤ a\ = 0.90.

Solution:

First note that since X台 ~ N (0, 1), X台(2) ~ x) and they are random sample, 台(A)=1 X台(2) ~ x) . That is 台(3)=1 X台(2) ~ x) and 台(9)=4 X台(2) ~ x) . Further, if W1 ~ xl ) and W2 ~ x2 ) and W1 and W2 are

independent, then w(w)2(l)之(之)l(l)2(l) ~ Fll 户l2 . Hence, the probability becomes

P ╱ ≤ a\ = P ╱ /(/)6(3) ≤ a(6/3)\ = P (F3 户6 ≤ 2a) = 0.9

This implies that 2a = F3 户6 户0 ]9 . From the F table or R, we have F3 户6 户0 ]9 = 3.289. Hence a = 1.644.

Q2: Let denote the mean of a random sample of size 100 from a χ2 distribution with 50 degrees of freedom. Compute an approximate value for P (49 < < 52).

Solution:

Recall that if X台 ~ xl(2), then µ = E(X台 ) = ν and σ 2 = V (X台 ) = 2ν. Hence here µ = 50 and σ 2 = 100. Further, by the CLT, s N (µ, ) when n is suﬃciently large. That is, -′ s N (0, 1). Hence an

approximate value for the following probability is

49 _ 50 _ µ 52 _ 50

10/10 σ/′n 10/10

Q3: Suppose X1 , X2 , . . . , XA are a random sample from a population with probability density function

f (x) = exp ╱ _ 、 , _o < x < o.

Show that the estimator

= |X台 |

is unbiased for σ and ﬁnd the variance of the estimator.

Solution:

We ﬁrst ﬁnd the distribution of W台 = |X台 | and show that it has an Exponential distribution. To this end, we use the method of distribution function. The CDF of W台 is deﬁned by

Fw乞 (w) = P (W台 ≤ w) = P (|X台 | ≤ w) = P (_w ≤ X台 ≤ w) = P (X台 ≤ w)_P (X台 ≤ _w) = Fx乞 (w)_Fx乞 (_w) Since X台 is a continuous random variable, so is W台 and that the support of W台 is (0, o). The probability

density function of W台 , fw乞 (w), is obtained by taking derivative of Fw乞 (w) with respect to w which becomes

fw乞 (w) = fx乞 (w) + fx乞 (_w) = exp ╱ _ 、 + exp ╱ _ 、 = exp /_ 、, w > 0

That is, W台 = |X台 | ~ Exp(σ). Recall that if W台 ~ Exp(σ), E(W台 ) = σ and V (W台 ) = σ 2 . Hence

E() = E( |X台 |) = E(|X台 |) = σ

This means is an unbiased estimator of σ. Further V () = V ( |X台 |) = V (|xA乞 |) = .

Second Method:

We can ﬁnd E(|X台 |) and V (|X台 |) directly as follows.

E(|X台 exp ╱ _ 、dx = -o(0) _x exp / 、dx + 0 o x exp /_ 、dx

2 2

For the variance, we ﬁrst ﬁnd E(|X台 |2 ) as follows.

E(|X台 |2 ) = -o(o) |x|2 exp ╱ _ 、dx = -o(0) x2 exp / 、dx + 0 o x2 exp /_ 、dx

= σ 2 + σ 2 = 2σ2

Hence, V (|X台 |) = E(|X台 |2 ) _ [E(|X台 |)]2 = σ 2 . The rest is the same as before. i.e.,

E() = E( |X台 |) = E(|X台 |) = σ

This means is an unbiased estimator of σ. Further V () = V ( |X台 |) = V (|xA乞 |) = .

Q4: Nineteen people volunteer to participate in a memory experiment. The volunteers are randomly divided into two groups. Group 1 (10 volunteers) will be given a list of words to study and Group 2 (9 volunteers) will listen to a reading of the same list of words. After half an hour, the subjects will be asked to write down all words they can remember. The scores for each of the subjects are shown below:

Group 1 Group 2

We would like to determine whether people’s memory diﬀers depending on whether they have read something or heard something. Assuming the scores follow a normal distribution for both groups and that the population variances are equal, ﬁnd a 95% conﬁdence interval for the diﬀerence in mean scores of the two methods and and provide an interpretation for the interval.

Solution:

Recall that under the assumptions that two populations are independent and have (approximately) normal distribution with equal variances, a 100(1 _ α)% C.I for the diﬀerence of their means (µ1 _ µ2 ) based on the

t distribution is given by

1 _ 2 呈 tAl +A2 -2 户1 - ←Ss(2)( + )

Where Ss(2) is the pooled sample variance obtained by Sp(2) = (Al -1)Al(扌)1)扌2(2) .

Here, we have n1 = 10, n2 = 9, and obtain 1 = 49, 2 = 38. Further s1(2) = 119.556, s2(2) = 114 which give ss(2) = 116.941. Also, from the t Table, we have t17 户0 ]975 = 2.1098. Hence a 95% C.I for µ 1 _ µ2 becomes

y¯1 _ y¯2 呈 t17 户0 ]975 ← 116.941( + ) = (0.517, 21.483)

Note that since the resulting C.I does not contain 0, there is evidence that µ 1 µ2 with 95% conﬁdence level.

Q5: A vending machine at the local ice rink dispenses hot chocolate, among other things. Ideally, the amount dispensed should not vary a lot (since otherwise, those will too little will not be happy and also those who get too much may spill!). To examine the variation, a random sample of 20 cups of hot chocolate dispensed by such a machine are examined. Their average amount dispensed was 174.1 ml and their variance was 8.8 ml2 . Find a 95% conﬁdence interval for the true variance of amount dispensed.

Solution:

Recall that a 100(1 _ α)% C.I for the population variance σ 2 is

╱ 11户之2(S2) , n2A-1(_)户之(2)2 \

Here we have n = 20 sample, = 174.1, s2 = 8.8, and 1 _ α = 0.95 or α = 0.05. Further, using the Chi-square table, we have x1(2)9 户0 ]025 = 32.852 and x1(2)9 户0 ]975 = 8.906. Substituting these values, a 95% C.I for

σ 2 becomes

90(女)6(8.)8 、= (5.089, 18.774)

Q6: Suppose that Y1 , ..., YA are a random sample from Gamma distribution with parameter (α = 2, β). Show that is a consistent estimator for β 2 .

Solution:

Recall that if W ~ Gamma(α, β), then µ = E(W) = αβ . Hence, for this question, µ = E(Y台 ) = 2β. By LLN, we know that is a consistent estimator of µ which is 2β .

Further, recall that if θˆ is a consistent estimator of θ and g is a continuous function, then g(θˆ) is a consistent estimator of g(θ). Because g(x) = x2 is a continuous function, 2 is a consistent estimator of 4β2 , and similarly, is a consistent estimator of β2 .