Stat Assignment 10A XXXXXXXXXXA researcher for the Goodyear tire company wants to know if the ages of cars are equally distributed among the three categories shown below for a sample of 30 car owners....

Stat Quiz 1B
Stat Assignment 10A
1) A researcher for the Goodyear tire company wants to know if the ages of cars are equally
distributed among the three categories shown below for a sample of 30 car owners. At Alpha = .05, can
it be considered that the ages are equally distributed? Use the chi-square goodness of fit procedure
Answer: DATA MISSING
2) The advisor of the college science club believes that the club membership is 10% freshmen, 20%
sophomores, 40% juniors, and 30% seniors. This year’s membership consists of 14, 19, 51, and 16,
respectively for the four groups. At Alpha = .10, test the advisors claim.
Answer: Here total number of observations = 14+19+51+16 = 100
Therefore, expected frequencies for freshmen = 10, sophomores = 20, juniors = 40 and seniors = 30.
Hence, the observed value o f the chi square goodness of fit statistic, χ2 = (10-14)2/10 + (20-19)2/20 +
(40-51)2/40 + (30-16)2/30 = 11.2083
Now, χ2 follows chi-square distribution with d.f. = 3 under null hypothesis
We reject the null hypothesis at 5% level of confidence if observed χ2 > χ20.05;3 = 7.82
Clearly, here observed χ2 > χ20.05;3. Therefore, at 5% level of confidence we reject the null hypothesis and
conclude that the claim is not significant.
Stat Assignment 11A
1) An economist believes that the median cost per pound of beef “on-the-hoof” is $5.00. A sample
of 22 buyers is shown below (in $ per pound.) At Alpha = .10, test the economist’s claim.
5.35 5.16 4.97 4.83 5.05 5.19
4.78 4.93 4.86 5.00 4.63 5.06
5.19 5.00 5.05 5.10 5.16 5.25
5.16 5.42 5.13 5.27

Answer: If median cost per pound of beef is $5 then proportion of costs under $5 is 0.5.
We define, p = proportion of costs less than or equal to $5 in the population.
Here, we are to test H: p = 0.5 vs. K: p ≠ 0.5
Here, the observed proportion, p = 8/22 = 0.36
The test statistic is, Z = (p – 0.5)/√[p(1-p)/n) which follows N(0,1) distribution under H. Here, sample size,
n = 22, observed p = 0.36. Therefore, observed Z = -1.368. We reject H at alpha = 0.1 if |observed
Z|>Z0.05 = 1.645. Clearly, |observed Z|<1.645. Therefore, we accept H at alpha = 0.1 and conclude that
the economist’s claim is valid.
2) A researcher claims that the median age of viewer’s of the Daily Show is 39. 75 viewer’s were
surveyed and 27 were under age 39. At Alpha = .02, test the claim.
Answer: If median age of viewer’s of the Daily Show is 39 then proportion of viewers under age 39 = 0.5.
Let, p = proportion of viewers under age 39
Here, we are to test H: p = 0.5 vs. K: p ≠ 0.5
Here, the observed proportion, p = 27/75 = 0.36
The test statistic is, Z = (p – 0.5)/√[p(1-p)/n) which follows N(0,1) distribution under H. Here, sample size,
n = 75, observed p = 0.36. Therefore, observed Z = -2.526. We reject H at alpha = 0.1 if |observed
Z|>Z0.01 = 1.282. Clearly, |observed Z|>1.282. Therefore, we reject H at alpha = 0.02 and conclude that
the researcher’s claim is not valid.
3) A statistician claims that a study skills course will improve student performance in the area of
reasoning skills. Nine students were pre-tested, they then took the study skills course, followed
by a post-test as indicated in the table below. At alpha = .05, did the course improve their
reasoning skills?
1 2 3 4 5 6 7 8 9
Pr 80 76 74 83 92 78 91 74 88
82 78 73 85 95 79 93 78 88
Student
etest
Postest

Answer: Here we will use paired sample t-test. We assume that the pretest scores follows Normal
distribution with mean µ1 and the postest scores follows Normal distribution with mean µ2. Here we are
to test, H: µ1 = µ2 Vs. K: µ2 > µ1
The test results are given below:
t-Test: Paired Two Sample for Means
Postest Pretest
Mean 83.44444 81.77778
Variance 54.77778 50.19444
Observations 9 9
Pearson Correlation 0.9795

Hypothesized Mean
Difference 0
df 8
t Stat 3.333333
P(T<=t) one-tail 0.005167
t Critical one-tail 1.859548
P(T<=t) two-tail...