1. Classify as independent or dependent samples: The number of crimes committed in a city before and after a new neighborhood patrol program is instituted. (Points :2) independent dependent

Dependent

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 20 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 4 grams. The normally-distributed weights (in grams) of these pennies are as follows: 5, 5, 9, 2, 7, 1, 7, 8, 3, 4. 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? (Points :10)

Here the null hypothesis is Ho: μ ≤ 4 and the alternative hypothesis is Ha: μ > 4.

Sample mean, xbar = 5.1

Sample standard deviation, s = 2.6437

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 = (5.1 - 4)/(2.6437/√10) = 1.3158

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 =1.3158 1.383

So we fail to reject the null hypothesis Ho.

Thus the mean weight of each penny is not at least 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? (Points :10)

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 11 consumers two choices: Do they prefer Coke or Pepsi? Exactly 9 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? (Points :10)

Here the null hypothesis is Ho: p = 0.70 and the alternative hypothesis is Ha: p ≠ 0.70.

The sample proportion, pbar = x/n = 9/11 = 0.8182

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 = (0.8182– 0.70)/Sqrt(0.70*0.30/11) = 0.8553

Since a = 0.02, the critical value is given by,

Critical value = ±2.33.

The decision rule is Reject Ho if |z| > 2.33

or Reject Ho if z < -2.33 or z > 2.33.

Here, |z| = 0.8553 < 2.33

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 second group is

xbar2 = 15.01 and 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. Thus the means of the two populations are not equal