Download 1 - JustAnswer

1. Classify as independent or dependent samples: The prices of a dozen watches, and the
prices of a dozen cars. (Points :2) independent dependent
Independent
2. Classify as independent or dependent samples: The rainfall in the US in two
consecutive years. (Points :2) independent dependent
Dependent
3. The pulse rate of a population of 20 volunteers is tested before and after exercise. In
each case, the pulse rates are normally distributed. Can the z test be used to evaluate the
difference in the means of each sample? (Points :3) Yes No
No. A paired t-test is required.
4. The US Mint selects ten pennies from the production line to test the hypothesis that the
mean weight of each penny is 4 grams. The normally-distributed weights (in grams) of
these pennies are as follows: 7, 6, 1, 6, 7, 8, 8, 9, 10, 7. Assume = 0.05. · 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 = 4 and the alternative hypothesis is Ha: μ ≠ 4.
Sample mean, xbar = 6.9
Sample standard deviation, s = 2.4244
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 = (6.9 - 4)/(2.4244/√10) = 3.7826
Since a = 0.05, from Student’s t distribution with (n-1) = 9 degrees of freedom the critical
value is given by,
Critical value = ±2.262
The decision rule is Reject Ho if |t| > 2.262
Or Reject Ho if t < -2.262 or t > 2.262.
Here, |t| = 3.7826 > 2.262
So we reject the null hypothesis Ho.
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 8, 9, 4, 6, 9, 6, 3, 1, 10, 3, 7, 10, 7, 5, and 6 grams. Assume = 0.10. · 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? (Points :10)
Here the null hypothesis is Ho: σ ≤ 2.0 and the alternative hypothesis is Ho: σ > 2.0
The sample standard deviation, s = 2.7115
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)( 2.71152)/22 = 14*7.3524/4 = 25.7333
Since a =0.10, from the Chi-square distribution with (n-1) = 14 degrees of freedom the
critical value is given by,
Critical value = 21.064
The decision rule is Reject Ho if χ2 > 21.064.
Here, χ2 = 25.7333 > 21.064.
So we reject the null hypothesis Ho.
Thus the standard deviation in their weights must be greater than 2.0 grams.
6. A telephone survey gives 84 consumers two choices: Do they prefer Coke or Pepsi?
Exactly 58 of those surveyed state that they prefer Coke. Assuming that = 0.05, test the
hypothesis that the proportion of the population that prefers Coke is 60%. · 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? (Points :10)
Here the null hypothesis is Ho: p = 0.60 and the alternative hypothesis is Ha: p ≠ 0.60.
The sample proportion, pbar = x/n = 58/84 = 0.6905
The test statistic for testing Ho is given by,
z = (pbar – 0.60)/Sqrt(0.60*0.40/n), follows Standard Normal distribution
Thus the test statistic,
z = (0.6905 – 0.60)/Sqrt(0.60*0.40/84) = 1.6927
Since a = 0.05, the critical value is given by,
Critical value = ±1.96.
The decision rule is Reject Ho if |z| > 1.96
Or Reject Ho if z < -1.96 or z > 1.96.
Here, |z| = 1.6927 < 1.96
So we fail to 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: 13.7 14.2 13.2 11.3 12.6 10.0 10.1 14.1 13.8 13.2 The times
required by sprinters in the second group are as follows: 17.3 10.6 14.0 15.9 15.1 13.4
18.9 17.2 17.6 10.1 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? (Points
:10)
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 = 12.62 and s1 = 1.5943
The mean and standard deviation of the first group is
xbar2 = 15.01and s2 = 2.9786
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*1.5943^2 + 9*2.9786^2)/18] = 2.3889
Thus the test statistic is given by,
t = (12.62 – 15.01)/ {2.3889*Sqrt[(1/10 + 1/10)]} = -2.2371
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| = 2.2371 > 2.101
So we reject the null hypothesis Ho.