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COLUMBIA BUSINESS SCHOOL
Fall 2001 Professor Paul Glasserman B6014: Managerial Statistics 403 Uris Hall
Solutions to Practice Final Exams #1 and
EXAM
a) Let μ be true average number of calories in the Diet-Burger. The null hypothesis,
H 0 , is μ = 400 and the alternative is μ >400. Therefore, the probability of a Type I
error (α) is just the probability of rejecting the null when it is true. If the null is true,
then μ = 400, so we need to calculate (recall n = 40 and σ=30):
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P X > = P Z > PZ
The probability of a Type I error is only 1.74%.
b) Answer: higher. With a sample of size 10, the likelihood of making a Type I error increases since the standard error increases and thus we are more likely to get a sample mean that deviates from the population mean by 10 units.
c) Assume μ=422. Then the null hypothesis is FALSE. A Type II error occurs when a false null is not rejected (i.e., is accepted). This occurs when the sample mean is less than 410. So we must calculate (recall n = 40 and σ = 30):
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P X < = P Z < PZ
The probability of a Type II error is only 0.57%.
2. This is a hypothesis test. Let p be the proportion of coupons that are redeemed. We set the null hypothesis to be that there has not been any change in people coupon usage habits.
I.e., H 0 : p = 0.06 and HA : p ≠ 0.06. The sample proportion is p ˆ^ = 315/5431 = 5.80%. Let's
calculate the p -value.
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P p < = P Z < PZ
The p -value is double this number, so p -value=2(0.2676) = 53.52%. This is a rather high p -value that suggests that there is little evidence against the null hypothesis. (Put another way, if p =0.06 so that 6% of coupons are redeemed on average, then there is a 53.52% chance of observing a sample that deviates from 6% by more than the amount
we observed. So this suggests that a sample like the one we have is very likely to occur by chance, even if the population proportion is 6%.) The data do not suggest that the proportion of responses is different from 6%.
3. The sample proportion of failures is 12/50 = 0.24. The 99% confidence interval is:
a) (A) t -stat for TVAD : 0.127/0.017=7. (B) p -value for TVAD : almost 0% (C) R-squared: SSR/SST=(1.971)/(1.971+0.447)=0.815.
(D) Standard Error of Estimate: s (^) e =
n − K − 1
SSE
(E) Degrees of Freedom, Error: n − K −1= n − 3 −1=18.
b) If coupons are given out, then we estimate the average increase in sales is 0.1 units, or $100,000. A 95% confidence interval is: 0.1 ± (2.101)(0.075) = 0.1 ± 0.158 = (−0.058,0.258). The expected effect of coupons on sales is somewhere between a decrease of $58, and an increase of $258,000.
c) Predicted sales are Y =0.376+0.127(4.7)+0.016(25)+0=1.3729 or $1,372,900.
d) For $20,000, the impact on sales of TVAD is (2)(0.127)=0.254, for NEWSAD is (20)(0.016)=0.32 and for coupons it is 0.10. Therefore the most cost-effective medium is NEWSAD. Expected sales increase is $320,000.
e) The t -stats for the three media are: for TVAD it is 0.127/0.017=7.47, for NEWSAD it is 0.016/0.003=5.33, and for COUPONS it is 0.100/0.075=1.33. Of these, coupons has the lowest t -stat (in absolute value), with a corresponding p- value of 20% (see the t - table with 18 degrees of freedom and double it). This medium may not have a significant effect on sales.
f) The test statistic is
=2.0 and the critical value from the t -table for 18
degrees of freedom is 1.734. Therefore we reject H 0. At the 5% level, we conclude that the effect of every $1,000 of NEWSAD on sales is most likely greater than $10,000.
g) Note that the R^2 is high at 81.5%, and two of the variables have a very significant relationship with sales. However, with an se of 0.1575, any prediction of sales will be accurate to about plus or minus 2(0.1575)=0.315 or $315,000, which limits the utility of the model.
the null. There are two possibilities, if we rejected because Z 0 < −zα/2, then when n 1 becomes n 2 , (if everything else stays the same) the statistic get even smaller (more negative) so we still reject. A similar argument can be made for the other side (when Z 0 > zα/2). But we have not taken into account the sample standard deviation s (^) X. The problem does not state whether this increases or decreases, so it is possible that we can end up accepting the null hypothesis (if for example the standard deviation increases significantly).
a) Answer: A 95% CI for μ is: [ 4.5−1.96(1) , 4.5+1.96(1) ] = (2.54,6.46).
b) FALSE: The p -value is not the probability that the null hypothesis is true. The p - value is the smallest significance level at which we reject the null. FALSE: The p -value is not the probability of a Type I error. The significance level (α) is the probability of a Type I error. TRUE: A small p -value indicates strong evidence in favor of the alternative hypothesis. FALSE: This is almost true, but we do not like to say that the test provides strong evidence for the null hypothesis. The testing procedure can only result in two possibilities: i) Strong evidence against the null hypothesis. ii) Not strong evidence against the null hypothesis.
2. a)
(A) R-squared = 74. 9 %.
(B) Degrees of freedom for Regression = K = 4. (C) Degrees of freedom for Error = n − K −1 = 46− 4 −1 = 41. (D) SSR = 5444.3−1364.8 = 4079.5.
b) The interval is found by simply considering a 95% confidence interval on β 4 :
Therefore in this case, men make anywhere from 5.00($1000) = $5,000 more than women to 2.07($1000)=$2,070 less than women. The significance of this is
provided by the p -value of the β 4 variable, it is found by first calculating a t -stat:
1.468/1.804=0.813. Now look up 0.813 in the z -table, and we get 0.2910. Therefore the p -value (for a two-sided test) is 2(0.5-0.2910)=41.8%: not much evidence at all. The difference is not significant. (This could also have been seen from the fact that the confidence interval includes 0.)
c) One year of education is worth anywhere from 1.904 ± (1.96)(0.3869) = 1.904 ± 0.758 = (1.146,2.662). This is with 95% confidence.
d) We want to test: H 0 : β 1 = 0.950 versus HA : β 1 ≠ 0.950 at the 1% level. Note since the
test is not versus the value 0, we cannot use the p -value or t -stat directly from the
output. We calculate the t -stat to be 2. 03.
This falls within the
thresholds of ±2.575, therefore we accept the null hypothesis. It is still likely that each year of employment at Alphatech increases your salary by $950. The p -value of this test is: 2(0.5-0.4788)=4.24%.
e) Plugging in the numbers: SALARY = 24.6 + 0.656(4)- 0.115(0) + 1.90(9) - 1.47(1)=42.854, or $42,854. A 95% confidence interval for the average salary of a person with these attributes would be:
42.854 ± (1.96)
that is, somewhere between $41,171 and $44,537.
a) A 99% confidence interval for μ=the true average summer income is given by:
33.1 ± 2.
b) Note that the interval from 31.5 to 33.8 is not symmetric around the sample mean 33.1. Therefore, we must solve this problem in two parts: first determine the content in the lower part of the interval (from 31.5 to 33.1) and then the content in the upper part of the interval (from 33.1 to 33.8). The standard error is: 5.0/√45 = 0.75. For the lower part,
P(31.5 < X < 33.1) =
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P Z =P(−2.133 < Z < 0 ) = 0.4834.
For the upper part,
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P 0 Z =P(0 < Z < 0.933) = 0.3238.
So, the probability content of the interval is 0.4834 + 0.3238 = 0.8072 = 80.72%. 4. a)
(A) Multiple R = R − squared = 0. 969 = 0. 984.
(B) Degrees of freedom for Regression = K = 3. (C) SSR = 29,657.8−916.7 = 28,741.1. (D) Degrees of freedom for Residual = n − K −1 = 14− 3 −1=10.
b) The test is: H 0 : β 3 = 0 vs. HA : β 3 ≠ 0. At the 5% level, we calculate our t -stat, it is
0.0859/0.0379=2.266. So, using the t -table with n − K −1=10 degrees of freedom, the threshold values are ±2.228. Therefore we are (slightly) outside the acceptance
region and therefore REJECT the null hypothesis (that β 3 =0) and conclude that
disposable income does have an effect on sales.
c) Each unit of disposable income has what effect on sales of Crest? With 95% confidence we are sure it is somewhere between 0.0859 ± 2.228(0.0379) = (0.00146,0.17034) or somewhere between $1.46 million and $170.34 million. Therefore an increase in $
