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1. Classify as independent or dependent samples: The prices of a dozen watches, and the
prices of a dozen cars. Independent dependent
Independent
2. Classify as independent or dependent samples: The effectiveness of Tylenol on twenty
patients, and the effectiveness of aspirin on another twenty patients. 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? Yes No
Yes, usually when n1 and n2≥30, values of population standard deviations can be
approximated by s1 and s2
4. The US Mint selects ten pennies from the production line to test the hypothesis that the
mean weight of each penny is at most 4 grams. The normally-distributed weights (in
grams) of these pennies are as follows: 4, 10, 10, 5, 4, 10, 6, 4, 9, 6. Assume ? = 0.10.
•?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?
Here the null hypothesis is Ho: μ ≤ 4 and the alternative hypothesis is Ha: μ > 4.
Sample mean, xbar = 6.8
Sample standard deviation, s = 2.6583
The test statistic for testing Ho is given by,
t = (xbar - 4)/(s/√n), follows a Student’s t distribution with (n-1) d.f..
Thus the test statistic
t = (6.8 - 4)/( 2.6583/√10) = 3.3308
Since a = 0.10, from Student’s t distribution with (n-1) = 9 degrees of freedom the critical
value is given by,
Critical value = 1.383
The decision rule is Reject Ho if t > 1.383
Here, t = 3.3308 > 1.383
So we reject the null hypothesis Ho. Thus the mean weight of each penny is not at most
4 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 3.0 grams.
Fifteen watch springs are selected from the production line and measured; their weights
are 2, 4, 10, 10, 8, 8, 9, 10, 9, 5, 3, 2, 4, 2, and 2 grams. Assume ? = 0.01. •?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 ≤ 3.0 and the alternative hypothesis is Ho: σ > 3.0
The sample standard deviation, s = 3.3352
The test statistic for testing Ho is given by,
χ2 = (n-1)s2/32, follows a Chi-square distribution with (n-1) degrees of freedom.
Thus the test statistic
χ2 = (15 -1)( 3.33522)/32 = 17.3037
Since a =0.01, from the Chi-square distribution with (n-1) = 14 degrees of freedom the
critical value is given by,
Critical value = 29.141
The decision rule is Reject Ho if χ2 > 29.141.
Here, χ2 = 17.3037 < 29.141.
So we fail to reject the null hypothesis Ho.
Thus the standard deviation in their weights must not be greater than 3.0 grams.
6. A telephone survey gives 601 consumers two choices: Do they prefer Coke or Pepsi?
Exactly 248 of those surveyed state that they prefer Coke. Assuming that ? = 0.02, test
the hypothesis that the proportion of the population that prefers Coke is 70%. •?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 = 0.70 and the alternative hypothesis is Ha: p ≠ 0.70.
The sample proportion, pbar = x/n = 248/601 = 0.4126
The test statistic for testing Ho is given by,
z = (pbar – 0.70)/Sqrt(0.70*0.30/n), follows Standard Normal distribution
Thus the test statistic,
z = (0.4126 – 0.70)/Sqrt(0.70*0.70/601) = - 15.3725
Since a = 0.02, the critical value is given by,
Critical value = ±2.326.
The decision rule is Reject Ho if |z| > 2.326
Or Reject Ho if z < -2.326 or z > 2.326.
Here, |z| = 15.3725 > 2.326
So we reject the null hypothesis Ho.
7. Two groups of ten sprinters run 100 meters. The times required by sprinters in the first
group are as follows: 10.2 12.3 12.5 10.4 11.1 11.5 10.8 11.8 13.4 10.0 The times
required by sprinters in the second group are as follows: 15.5 15.2 16.0 12.4 17.4 19.7
14.8 10.4 15.8 11.9 Assuming that ? = 0.05, 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 = 11.4 and s1 = 1.1075
The mean and standard deviation of the first group is
xbar2 = 14.91 and s2 = 2.7339
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*1.1075^2 + 9*2.7339^2)/18] = 2.0858
Thus the test statistic is given by,
t = (11.4 – 14.91)/ {2.0858*Sqrt[(1/10 + 1/10)]} = -3.7629
Since a = 0.05, from Student’s t distribution with (n1 + n2 -2) = 18 degrees of freedom the
critical value is given by,
Critical value = ±2.101
The decision rule is Reject Ho if |t| > 2.101
Or Reject Ho if t < -2.101 or t > 2.101.
Here, |t| = 3.7629 > 2.101
So we reject the null hypothesis Ho.