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1. Classify as independent or dependent samples: White blood cell counts for a group of
volunteers before being inoculated, and white blood cell counts for a second group of
volunteers after being inoculated. (Points :2) independent dependent
Independent
2. Classify as independent or dependent samples: Average selling price for ten homes in a
given neighborhood, and average selling price for another ten homes in that
neighborhood. (Points :2) independent dependent
Independent
3. An aerospace parts factory has two separate production lines making fasteners; each
production line can be regarded as creating a separate population of fasteners, whose
widths are normally distributed. A sample of 50 fasteners is selected from each
production line for quality control inspection. The width of each selected fastener is
measured; and the standard deviations s1 and s2 in the measured widths of the selected
fasteners from each production line are calculated. If the design standard deviation in
fastener width for each production line population is unknown, can the sample values s1
and s2 be used instead? (Points :3) Yes No
Yes.
4. The US Mint selects ten pennies from the production line to test the hypothesis that the
mean weight of each penny is at least 6 grams. The normally-distributed weights (in
grams) of these pennies are as follows: 6, 6, 8, 5, 9, 5, 9, 2, 3, 8. Assume = 0.01.
•?????????State the null and alternate hypotheses •?????????Calculate the sample mean
and standard deviation •?????????Determine which test statistic is appropriate (z or t),
and calculate its value. •?????????Determine the critical value(s). •?????????State your
decision: Should the null hypothesis be rejected? (Points :10)
Here the null hypothesis is Ho: μ ≤ 6 and the alternative hypothesis is Ha: μ > 6.
Sample mean, xbar = 6.1
Sample standard deviation, s = 2.4244
Here we use the t-test.
The test statistic for testing Ho is given by,
t = (xbar - 6)/(s/√n), follows a Student’s t distribution with (n-1) d.f..
Thus the test statistic
t = (6.1 - 6)/(2.4244/√10) = 0.1304
Since a = 0.01, from Student’s t distribution with (n-1) = 9 degrees of freedom the critical
value is given by,
Critical value = 2.821
The decision rule is Reject Ho if t > 2.821
Here, t = 0.1304 < 2.821
So we fail to reject the null hypothesis Ho.
Thus we cannot conclude that the mean weight of each penny is at least 6 grams.
5. A watch manufacturer creates watch springs whose properties must be consistent. In
particular, the standard deviation in their weights must be no greater than 2.0 grams.
Fifteen watch springs are selected from the production line and measured; their weights
are 10, 4, 1, 6, 9, 8, 5, 6, 1, 2, 6, 1, 10, 2, and 5 grams. Assume a = 0.05. •?????????State
the null and alternate hypotheses •?????????Calculate the sample standard deviation
•?????????Determine which test statistic is appropriate (chi-square or F), and calculate its
value. •?????????Determine the critical value(s). •?????????State your decision: Should
the null hypothesis be rejected?
Here the null hypothesis is Ho: σ ≤ 2.0 and the alternative hypothesis is Ho: σ > 2.0
The sample standard deviation, s = 3.2175
Here we use the Chi-square test statistic.
The test statistic for testing Ho is given by,
χ2 = (n-1)s2/22, follows a Chi-square distribution with (n-1) degrees of freedom.
Thus the test statistic
χ2 = (15 -1)(3.21752)/22 = 36.2333
Since a = 0.05, from the Chi-square distribution with (n-1) = 14 degrees of freedom the
critical value is given by,
Critical value = 23.685
The decision rule is Reject Ho if χ2 > 23.685.
Here, χ2 = 36.2333 > 23.685.
So we reject the null hypothesis Ho.
Thus we cannot conclude that the standard deviation in their weights must be no greater
than 2.0 grams.
6. A telephone survey gives 367 consumers two choices: Do they prefer Coke or Pepsi?
Exactly 266 of those surveyed state that they prefer Coke. Assuming that a = 0.01, test
the hypothesis that the proportion of the population that prefers Coke is 30%.
•?????????State the null and alternate hypotheses •?????????Calculate the sample
proportion •?????????Calculate the value of the test statistic. •?????????Determine the
critical value(s). •?????????State your decision: Should the null hypothesis be rejected?
Here the null hypothesis is Ho: p = 0.30 and the alternative hypothesis is Ha: p ≠ 0.30.
The sample proportion, pbar = x/n = 266/367 = 0.7248
The test statistic for testing Ho is given by,
z = (pbar – 0.30)/Sqrt(0.30*0.70/n), follows Standard Normal distribution
Thus the test statistic,
z = (0.7248 – 0.30)/Sqrt(0.30*0.70/367) = 17.7584
Since a = 0.01, the critical value is given by,
Critical value = ±2.576.
The decision rule is Reject Ho if |z| > 2.576
Here, |z| = 17.7584 > 2.576
So we reject the null hypothesis Ho.
Thus we cannot conclude that the proportion of the population that prefers Coke is 30%.
7. Two groups of ten sprinters run 100 meters. The times required by sprinters in the first
group are as follows: 13.4 10.4 14.9 10.1 12.5 13.8 13.7 11.1 11.0 10.9 The times
required by sprinters in the second group are as follows: 12.1 15.1 12.1 11.4 16.7 18.1
16.3 10.1 18.1 16.9 Assuming that a = 0.10, test the hypothesis that the means of the two
populations are equal. •?????????State the null and alternate hypotheses
•?????????Calculate the mean and standard deviation for each group •?????????Calculate
the value of the test statistic. •?????????Determine the critical value(s). •?????????State
your decision: Should the null hypothesis be rejected?
Here the null hypothesis is Ho: μ1 = μ2 and the alternative hypothesis is Ha: μ1 ≠ μ2.
The mean and standard deviation of the first group is
xbar1 = 12.18 and s1 = 1.6871
The mean and standard deviation of the second group is
xbar2 = 14.69 and s2 = 2.9861
The test statistic for testing Ho is given by,
t = (xbar1 – xbar2)/{sp*Sqrt[(1/n1)+(1/n2)]} follows a Student’s t distribution with
(n1 + n2 -2) degrees of freedom, where sp = Sqrt{[(n1-1)s1^2+(n2-1)s2^2]/ (n1 + n2 -2)] is
the pooled standard deviation.
Here, sp = Sqrt[(9*1.6871^2 + 9*2.9861^2)/18] = 2.4252
Thus the test statistic is given by,
t = (12.18 – 14.69)/ {2.4252*Sqrt[(1/10 + 1/10)]} = -2.3143
Since a = 0.10, from Student’s t distribution with (n1 + n2 -2) = 18 degrees of freedom the
critical value is given by,
Critical value = ±1.734
The decision rule is Reject Ho if |t| > 1.734
Here, |t| = 2.3143 > 1.734
So we reject the null hypothesis Ho. Thus the means of the two populations are not equal.