Statistics Sample Exam 4
Chapters 8 & 9: Hypothesis Testing & Inferences from Two Samples
1. A test of hypothesis is always about a population parameter.
2. The observed value of a test statistic is the value calculated for a sample statistic.
ex) z scores, t distribution, chi-square(X2)
3. As the sample size gets larger, both type I and type II errors decrease.
4. Define type I error.
Rejecting a hypothesis when it is true (false positive)
5. Define type II error.
Failing to reject a hypothesis when it is false (false negative)
6. The value of α is called the probability of type I error
7. The value of β is called the probability of type II error
8. The value of 1-β is called the correct decision ( probability of rejecting a false
hypothesis)
9. In 1990, 5.8% of job applicants who were tested for drugs failed the test. At the
0.01 significance level, test the claim that the failure rate is now lower if a simple
random sample of 1520 current job applicants results in 58 failures. Does the
result suggest that fewer job applicants now use drugs?
given: p = 0.058, α = 0.01, n = 1520, x = 58
a. State the null and alternate hypothesis.
H0: p = 0.058
H1: p < 0.058 (claim, LTT)
b. Find the value of sample proportion.
p^ = x/n = 58/1520 = 0.03816
c. Calculate the value of test statistic.
z = p^ - p / √(pq/n)
z = 0.03816 – 0.058 / √(.058 x 0.942 / 1520)
z = -3.3096
d. Find the critical value.
z = -2.325
e. Make a decision.
Reject H0 because the value of test statistic is inside the rejection region
Claim is true
There is sufficient evidence to support the claim that the failure rate
among job applicants for drugs is now lower.
Result suggests that fewer job applicants now use drugs.
f. Find the p-value and make a decision. Is this decision in agreement with
the previous one?
P-value: 0.0009
0.0009 < 0.01 – reject null hypothesis
this decision is in agreement with the previous one.
g. What is the probability of making type I error?
0.01
10. A sample of 54 bears has a mean weight of 182.9 lb. Let’s assume that the
standard deviation of weights of all such bears is known to be 121.8 lb, at α = 0.1.
Is there enough evidence to support the claim that the population mean of all
such bear weights is less than 200 lb?
given: n = 54, x = 182.9 lb, σ = 121.8 lb, α = 0.1 μ = 200 lb
a. State the null and alternate hypothesis.
H0: μ = 200
H1: u < 200 (claim, LTT)
b. Find the critical value(s).
z = -1.285
c. Calculate the value of test statistic.
z = x – μ / σ/√n
z = 182.9 – 200 / 121.8/√54
z = -1.0317
d. Make a decision.
Fail to reject the null hypothesis because the value of test statistic is inside
the non-rejection region.
Claim is false.
There is not a sufficient evidence to support the claim that the population
mean of all bear weights is less than 200 lb.
11. Sixteen new textbooks in the college bookstore, had prices with a mean of $70.41
and a standard deviation of $19.70. Use a 0.05 significance level to test the claim
that the mean price of a textbook at this college is less than $75?
given: n = 16, x = $70.41, s = $19.70, α = 0.05, μ = $75
H0: μ = 75
H1: μ < 75 (claim, LTT)
df: n – 1 = 15
(cv): t = ± 1.753
(ts) t = x – μ / s/√n
t = 70.41 – 75 / 19.70/√16
t = -0.9320
Fail to reject the null hypothesis because the value of test statistic is inside the
non-rejection region.
Claim is false.
There is not a sufficient evidence to support the claim that the mean price of a
textbook at this college is less than $75.
12. Tests in the author’s past statistics classes have scores with a standard deviation
equal to 14.1. One of his current classes now has 27 test scores with a standard
deviation of 9.3. Use a 0.01 significance level to test the claim that this current
class has less variation than past classes. Does a lower standard deviation
suggest that the current class is doing better? Assume the population is normal.
given: σ = 14.1, n = 27, s = 9.3, α = 0.01
H0: σ = 14.1
H1: σ < 14.1 {claim, LTT}
-cv: df: 26, α = 0.01
X2 = 12.198
-ts
X2 = (n-1)s2 /σ2
X2 = 26 x 9.32/14.12
X2 = 11.3110
Reject null hypothesis because the value of test statistic is inside the rejection
region.
Claim is true.
There is sufficient evidence to support the claim that this current class has less
variation than past classes.
A lower standard deviation does not suggest that the current class is doing better.
It just means that the grade is more homogeneous (closer to the mean). Mean
value might be different.
13. Among 843 smoking employees of hospitals with the smoking ban, 56 quit
smoking in one year after the ban. Among 703 smoking employees from work
places without the smoking ban, 27 quit smoking in one year.
given: n1 = 843, x1 = 56, n2 = 703, x2 = 27
a. Is there a significant difference between the two proportions? Use a 0.01
significance level
p = 56 + 27/843 + 703 = 83/1546 = 0.0537
q = 1 – p = 0.9463
p1^ = x1/n1 = 56/843 = 0.0664
p2^ = x2/n2 = 27/703 = 0.0384
q1^ = 1- p1^ = 0.9336
q2^ = 1- p2^ = 0.9616
H0: p1 = p2
H1: p1 ≠ p2 (2TT)
critical values = ± 2.575
test statistic value: z = (p1^ - p2^) – (p1 – p2) /√((p q/n1) + (p q/n2))
z = (0.0664 – 0.0384) – 0 / √((0.0537 x 0.9463/843) + (0.0537x0.9463/703))
z = 2.4319
fail to reject null hypothesis
there is not a significant difference between the two proportions
b. Construct the 99% confidence interval for the difference between the two
proportions.
(p1^ - p2^) – E < p1 – p2 < (p1^ - p2^) + E
E = z α/2 √((p1^q1^/n1) + (p2^q2^/n2))
E = 2.575 √((0.0664 x 0.9336/843) + (0.0384 x 0.9616/703))
E = 0.0289
(0.0664 – 0.0384) – 0.0289 < p1 – p2 < (0.0664 – 0.0384) + 0.0289
-0.0009 < p1 – p2 < 0.0569
since 0 is included in this 99% confidence interval, there is no significant
difference between the two proportions
14. Company “A” claims that its yogurt cups contain, on average fewer calories than
that of a competitor. A sample of 50 such yogurt cups of company “A” produced
an average of 141 calories per cup with a standard deviation of 5.4 calories. A
sample of 40 yogurt cups of a rival company “B” produced an average of 144
calories per cup with a standard deviation of 6.3 calories.
given: n1 = 50, x1 = 141 calories, s1 = 5.4 calories, n2 = 40, x2 = 144 calories, s2 = 6.3
calories
a. Assuming that the calories of the yogurt cups for company “A” and
company “B” have different variances, use a 0.01 significance level to test
the claim.
H0: μ1 = μ2
H1: μ1 < μ2 (claim, LTT)
df = smaller of n1-1, n2-1 = 39
t = -2.426 (critical value)
(test statistic): t = (x1 – x2) – (μ1- μ2) / √((s12/n1) + (s22/n2))
t = 141 – 144/ √((5.42/50) + (6.32/40))
t = -2.3901
fail to reject null hypothesis
claim is false
there is not a sufficient evidence to support the claim that company A’s
yogurt cups contain, on average fewer calories than that of a competitor
b. Calculate the p-value for the test of previous part and make a decision. Is
decision in agreement with the previous one?
-2.426 < t = -2.3901 < -2.023
0.01 < p-value < 0.025
since p-value is greater than 0.01 (α) we fail to reject the null hypothesis
(therefore, the claim is false)
decision is in agreement with the previous one
c. Make the 98% confidence intervals for the difference between the two
means.
(x1 – x2) – E < μ1 – μ2 < (x1 – x2) + E, E = t α/2√(s12/n1 + s22/n2)
E = 2.426 √(5.42/50 + 6.32/40)
E = 3.0450
(141 – 144) – 3.0450 < μ1 – μ2 < (141-144) + 3.0450
-6.045 < μ1 – μ2 < 0.045
15. Assuming that the calories of the yogurt cups for company “A” and company “B”
have equal variances, repeat the previous question.
A: n = 50, x =141 calories, s = 5.4 calories
B: n = 40, x = 144 calories, x = 6.3 calories
a. H0: μ1 = μ2
H1: μ1 < μ2 (LTT, claim)
df = n1 + n2 – 2 = 50 + 40 – 2 = 88 ≈ 90 (when using table A-3)
α = 0.01
critical value: -2.368
test statistic value: t = (x1 – x2) – (μ1 – μ2) / √((s2p/n1) + (s2p/n2))
s2p = (n1 – 1)s12 + (n2 – 1)s22 / (n1 – 1) + (n2 – 1)
s2p = (50 – 1)5.42 + (40 – 1)6.32 / 49 + 39
s2p = 33.8267
t = (141 – 144) – 0 / √((33.8267/50) + (33.8267/40))
t = -2.4316
reject null hypothesis because the value of test statistic is inside the
rejection region
claim is true
there is sufficient evidence to support the claim that companyA’s yogurt
cups contain on average less calories than that of its competitors
b. -2.708 < t=-2.4316 < -2.426
0.0005 < p-value < 0.01
p-value less than 0,01, the result agrees with part a
c. (x1 – x2) – E < μ1 – μ2 < (x1 – x2) + E, E = t α/2√(sp2/n1 + sp2/n2)
E = 2.368 √((33.8267/50) + 33.8267/40))
E = 2.9216
(141 – 144) – 2.9216 < μ1 – μ2 < (141 – 144) + 2.9216
-5.9216 < μ1 – μ2 < -0.0784
16. Test a claim that weights of male college students have a larger variance than
female college students. Use a significance level of 0.05 and assume the
populations are normal.
males: n = 31, x = 168, s = 28
females: n = 29, x = 125, s = 25
given: α = 0.05, normal population
H0: σ12 = σ22
H1: σ12 > σ22 (claim, RTT)
df1: n1 – 1 = 30 (numerator degrees of freedom) df2: n2 – 1 = 28 (denominator
degrees of freedom)
critical value: F = 1.8687
test statistic value: F = s12/s22
F = 282/252 = 1.2544
we fail to reject the null hypothesis because the value of test statistic falls in the
non-rejection region
claim is false
there is not a sufficient evidence to support the claim that weights of male college
students have a larger variance than female college students