Stat - Questions

1.

Consider the following competing hypotheses and accompanying sample data drawn independently from normally distributed populations. Use Table 2.

H0: μ1 − μ2 ≥ 0

HA: μ1 − μ2 < 0

   = 249

 = 272

  s1 = 35

s2 = 23

  n1 = 10

n2 = 10

a-1.

Calculate the value of the test statistic under the assumption that the population variances are unknown but equal. (Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Test statistic

a-2.

Calculate the critical value at the 5% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

  Critical value

a-3.

Do you reject the null hypothesis at the 5% level?

Yes, since the value of the test statistic is not less than the critical value.

No, since the value of the test statistic is not less than the critical value.

Yes, since the value of the test statistic is less than the critical value.

No, since the value of the test statistic is less than the critical value.

b-1.

Calculate the value of the test statistic under the assumption that the population variances are unknown and are not equal. (Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Test statistic

b-2.

Calculate the critical value at the 5% level of significance. (Negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

  Critical value

b-3.

Do you reject the null hypothesis at the 5% level?

No, since the value of the test statistic is less than the critical value.

No, since the value of the test statistic is not less than the critical value.

Yes, since the value of the test statistic is less than the critical value.

Yes, since the value of the test statistic is not less than thecritical value.

2.

Consider the following sample data drawn independently from normally distributed populations with equal population variances. Use Table 2.

Sample 1

Sample 2

12.1

8.9

9.5

10.9

7.3

11.2

10.2

10.6

8.9

9.8

9.8

9.8

7.2

11.2

10.2

12.1

Click here for the Excel Data File

μ1 is the population mean for individuals with a CFA designation and μ2 is the population mean of individuals with MBAs

a.

Construct the relevant hypotheses to test if the mean of the second population is greater than the mean of the first population.

H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0

H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0

H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0

b-1.

Calculate the value of the test statistic. (Negative values should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places)

  Test statistic

b-2.

Calculate the critical value at the 1% level of significance. (Negative value should be indicated by a minus sign.Round your answer to 3 decimal places.)

  Critical value

b-3.

Do you reject the null hypothesis at the 1% level?

Yes, since the p-value is more than α

Yes, since the p-value is less than α.

No, since the p-value is less than α.

No, since the p-value is more than α

c.

Using the critical value approach, can we reject the null hypothesis at the 10% level?

No, since the value of the test statistic is not less than the critical value of -1.345.

Yes, since the value of the test statistic is less than the critical value of -1.345.

No, since the value of the test statistic is not less than the critical value of -1.761.

Yes, since the value of the test statistic is less than the critical value of -1.761.

3.

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a less than desirable battery life. They aim to compete with a new touch phone that is guaranteed to have a battery life more than two hours longer than the leading product. A recent sample of 120 units of the leading product provides a mean battery life of 5 hours and 40 minutes with a standard deviation of 30 minutes. A similar analysis of 100 units of the new product results in a mean battery life of 8 hours and 5 minutes and a standard deviation of 55 minutes. It is not reasonable to assume that the population variances of the two products are equal. Let new products and leading products represent population 1 and population 2, respectively. Use Table 2.

Sample 1 is from the population of new phones and Sample 2 is from the population of old phones. All times are converted into minutes.

a.

Set up the hypotheses to test if the new product has a battery life more than two hours longer than the leading product.

H0: μ1 − μ2 ≥ 120; HA: μ1 − μ2 < 120

H0: μ1 − μ2 ≤ 120; HA: μ1 − μ2 > 120

H0: μ1 − μ2 = 120; HA: μ1 − μ2 ≠ 120

b-1.

Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Test statistic

b-2.

Implement the test at the 5% significance level using the critical value approach.

Reject H0; the battery life of the new product is more than two hours longer than the leading product.

Do not reject H0; the battery life of the new product is more than two hours longer than the leading product.

Reject H0; the battery life of the new product is not more than two hours longer than the leading product.

Do not reject H0; the battery life of the new product is not more than two hours longer than the leading product.

4.

Consider the following competing hypotheses: Use Table 2.

H0: μD ≥ 0; HA: μD < 0

d-bar = −2.8, sD = 5.7, n = 12

The following results are obtained using matched samples from two normally distributed populations:

a.

At the 5% significance level, find the critical value(s). (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Critical value

b.

Calculate the value of the test statistic under the assumption that the difference is normally distributed.(Negative value should be indicated by a minus sign. Round your answer to 2 decimal places.)

  Test statistic

c.

What is the conclusion to the hypothesis test?

Reject H0 since the value of the test statistic is less than the critical value.

Do not reject H0 since the value of the test statistic is less than the critical value.

Do not reject H0 since the value of the test statistic is not less than the critical value.

Reject H0 since the value of the test statistic is not less than the critical value.

5.

Consider the following matched samples representing observations before and after an experiment. Assume that sample data are drawn from two normally distributed populations. Let the difference be defined as Before minus After. Use Table 2.

  Before

2.5

1.8

1.4

−2.9

1.2

−1.9

−3.1

2.5

  After

2.9

3.1

3.9

−1.8

0.2

0.6

−2.5

2.9

Click here for the Excel Data File

a.

Construct the competing hypotheses to determine if the experiment increases the magnitude of the observations.

H0: μD ≥ 0; HA: μD < 0

H0: μD = 0; HA: μD ≠ 0

H0: μD ≤ 0; HA: μD > 0

b-1.

Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Test statistic

b-2.

Approximate the p-value.

0.025 < p-value < 0.05

0.05 < p-value < 0.10

0.01 < p-value < 0.025

0.1 < p-value < 0.2

0.005 < p-value < 0.01

b-3.

Interpret the results at a 5% significance level.

Do not reject H0 since the p-value is less than α.

Do not reject H0 since the p-value is more than α.

Reject H0 since the p-value is less than α.

Reject H0 since the p-value is more than α.

c.

Do the results change if we implement the test at a 1% significance level?

Yes, Reject H0

Yes, Do not reject H0

No, Reject H0

No, Do not reject H0

6.

A diet center claims that it has the most effective weight loss program in the region. Its advertisements say, “Participants in our program lose more than 5 pounds within a month.” Six clients of this program are weighed on the first day of the diet and then one month later. Let the difference be defined as Weight on First Day of Diet minus Weight One Month Later. Use Table 2.

Client

Weight on First

 Day of Diet

Weight One

 Month Later

1

158

151

2

205

200

3

170

169

4

189

179

5

149

144

6

135

129

Click here for the Excel Data File

a.

Specify the null and alternative hypotheses that test the diet center’s claim.

H0: μD = 5; HA: μD ≠ 5

H0: μD ≤ 5; HA: μD > 5

H0: μD ≥ 5; HA: μD < 5

b.

Assuming that weight loss is normally distributed, calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

  Test statistic

c.

Approximate the p-value.

p-value < 0.005

p-value > 0.10

0.025 < p-value < 0.05

0.005 < p-value < 0.01

0.05 < p-value < 0.10

d.

At the 5% significance level, do the data support the diet center’s claim?

No since the p-value is more than α.

Yes since the p-value is more than α.

Yes since the p-value is less than α.

No since the p-value is less than α.

7.

(Use Excel) An engineer wants to determine the effectiveness of a safety program. He collects annual loss of hours due to accidents in 12 plants “before and after” the program was put into operation. Let the difference be defined as Before minus After. Use Table 2.

Plant

Before

After

Plant

Before

After

1

100

98

7

88

90

2

90

88

8

75

70

3

94

90

9

65

62

4

85

86

10

58

60

5

70

67

11

67

60

6

83

80

12

104

98

Click here for the Excel Data File

a.

Specify the competing hypotheses that determine whether the safety program was effective.

H0: μD ≤ 0; HA: μD > 0

H0: μD = 0; HA: μD ≠ 0

H0: μD ≥ 0; HA: μD < 0

b.

Assuming that hours are normally distributed, calculate the value of the test statistic. (Round your answer to 2 decimal places.)

  Test statistic

c.

At the 5% significance level, calculate the critical value. (Round your answer to 3 decimal places.)

  Critical value

d.

Is there sufficient evident to conclude that the safety program was effective?

Yes, since we reject H0.

Yes, since we do not reject H0.

No, since we reject H0.

No, since we do not reject H0.