Answers for Second Midterm Review Questions

Statistics 312 – Dr. Uebersax

Answers for second midterm review

Answers for Second Midterm Review Questions

1. What is the interpretation of a credible interval?

"A range in which one believes with x% certainty that a certain population

parameter falls."

2. What is the interpretation of a confidence interval?

In this course: an approximation to a credible interval.

3. When do you use a z- distribution and when do you use a t-distribution?

We use a z-distribution when we know the population standard deviation;

otherwise we use a t-distribution

4. In a sample of 25 cases, a variable has a mean of 11 and a standard deviation of 3. For a particular case in this sample, the variable has a value of 14.

(a) calculate a z score for this case: (should not be done; question dropped).

(b) calculate a tscore for this case:

5. What is the difference in shape between a z-distribution and a t-distribution?

The t-distribution has fatter tails.

6. Problem (single mean)

The average weight of a sample of 144 batteries is 2.3 oz., and the population standard deviation is 0.28 oz. Supply the upper and lower 90% limits of the CI in the population of all batteries. (Correction: also assume that the critical z value fora 90% CI is +/- 1.645.

, ,

7. Problem (two means, independent groups)

You collect sample data on how longelectronic devices manufactured at two different facilities work before failing. The results are as follows:

Plant 1: n1 = 100, sample mean = 115 hours, sample standard deviation = 8 hours.

Plant 2: n2= 150, sample mean = 125 hours, sample standard deviation = 10 hours.

You want to know if there is evidence of a difference in the quality of devices made in Plant 1 vs. Plant 2.

(a) What are the null and alternative hypothesis? (Assume a 2-tailed test.)

(b) Let α = 0.05. What does that mean?

"The probability of a Type 1 error (rejecting H0 when H0 is true) is 0.05."

Given H0, . So this drops from equation.

8. Problem (paired t-test)

You measure the strength of the same 5 steel rods (lbs. of weight supported) on a cold day and on a hot day. The results are below. Test whether or not the bearing capacity appears equal on hot and cold days. Extra columns are provided as workspace. (Use of these is optional, but recommended; not all may be necessary).

Rod / Cold day
(lbs.) / Hot day
(lbs.)
1 / 100 / 90
2 / 120 / 115
3 / 90 / 91
4 / 85 / 80
5 / 97 / 90

(a) What are the null and alternative hypotheses?(Assume a 2-tailed test.)

- or -

(where D = μ1 –μ2)

(b) What is the value of t?

Rod / Cold / Hot / D / D-Dbar / D-Dbar^2 / alpha / 0.05
1 / 100 / 90 / 10 / 4.8 / 23.04 / Mean(D) / 5.2
2 / 120 / 115 / 5 / -0.2 / 0.04 / n / 5
3 / 90 / 91 / -1 / -6.2 / 38.44 / df / 4
4 / 85 / 80 / 5 / -0.2 / 0.04 / std dev. D / 4.0249
5 / 97 / 90 / 7 / 1.8 / 3.24 / std. err. D / 1.8000
t / 2.8889

, , ,

df = number of pairs – 1 = 5 – 1 = 4 df.

9. Chi-squared goodness-of-fit test.

You predict that the weights of manufactured bolts follow a uniform distribution over the range of 1 oz. to 2 oz. You weigh 120 bolts and assign them to four bins as showed below. Perform a chi-squared test to see if the actual bolts follow the expected distribution. Extra columns are supplied for workspace.

Weight
(oz) / Obs. frequency
(O) / Predicted
probability
(π) / Expected
frequency
(E) / /
1.00–1.25 / 25 / 0.25 / 30 / -5 / 0.833
1.26–1.50 / 32 / 0.25 / 30 / 2 / 0.133
1.51–1.75 / 28 / 0.25 / 30 / -2 / 0.133
1.76–2.00 / 35 / 0.25 / 30 / 5 / 0.833

(a) What are the null and alternative hypotheses?

H0: The observed data come from a population characterized bythe predicted probabilitydistribution..

H1: Not H0.

(b) What is the value of the Pearson X2 statistic?

Ei = πin, n = 120, E = {30, 30, 30, 30}

Pearson= 0.833 + 0.133 + 0.133 + 0.833 = 1.933

df = (no. of categories – 1) = (4 – 1) = 3.

10. Chi-squared test of independence

Will not be on midterm.

11. Odds ratio

Customer Satisfaction
Design / Low / High
Old / 60 (= a) / 5 (= b)
New / 15 (= c) / 40 (= d)

(a) What is the value of the odds ratio (OR)?

ln(OR) = ln(32) = 3.4657

Given that the critical z values that define a 99% CI are +/- 2.576.

(b) What are the LL and UL for the log of the odds ratio?

LLln(OR) = 3.4657 – 2.576 (0.5553) = 2.035

ULln(OR) = 3.4657 + 2.576 (0.5553) = 4.896

exp(2.035) = 7.66, exp(4.896) = 133.76