# MATHEMATICS

**1. Use the paired t****–****interval procedure to obtain the required confidence interval for the mean difference. Assume that the conditions and assumptions for inference are satisfied.**

Ten families are randomly selected and their daily water usage (in gallons) before and after viewing a conservation video. Construct a 90% confidence interval for the mean of the difference of the “before” minus the “after” times if d (after-before) = -4.8 and Sd 5.2451

Before 33 33 38 33 35 35 40 40 40 31

After 34 28 25 28 35 33 31 28 35 33

- (1.5,8.1)
- (2.5,7.1)
- (1.8,7.8)
- (3.8,5.8)
- (2.1,7.5)

2.**From the sample statistics, find the value of the pooled estimate Pcap used.
**n

_{1 }= 36 n

_{2}= 418

x

_{1}= 7 x

_{2}= 132

Pcap =

3. **Provide an appropriate response.
**Do motivation levels between mid-level and upper-level managers differ? A randomly selected group of each were administered a survey, which measures motivation for upward mobility. The scores are summarized below:

Upper -Level | Mid-level | |

Sample size | 73 | 109 |

Mean score | 77.4 | 79.71 |

Standard Deviation | 10.6 | 6.43 |

Assuming equal population standard deviations, calculate the test statistic for determining whether the mean scores differ for upper-level and mid-level managers.

- -1.89c.-63.69 e. none of these
- -0.29 d. -1.74

4. **Use the paired t****–****interval procedure to obtain the required confidence interval for the mean difference. Assume that the conditions and assumptions for inference are satisfied.**

A test for abstract reasoning is given to a random sample of students before and after they complete a formal course in logic. The results are given below. Construct a 95% confidence interval for the difference in scores if d = after-before, Xbar_{d} = 3.7 and s_{d} = 4.945.

After 74 83 75 88 84 63 93 84 91 77

Before 73 77 70 77 74 67 95 83 84 75

- (-4.4, 11.8)
- (0.2, 7.2)
- (1.2,5.7)
- (1.0,6.4)
- (0.8,6.6)

5. **Find the appropriate test statistic/p****–****value.**

Do motivation levels between mid-level and upper-level managers differ? A randomly selected group of each were administered a survey, which measures motivation for upward mobility. The scores are summarized below:

Upper -Level | Mid-level | |

Sample size | 73 | 109 |

Mean score | 77.4 | 79.71 |

Standard Deviation | 10.6 | 6.43 |

Calculate the appropriate test statistic and give your conclusion for testing

Ho: Uj=Ua

Ha: Uj<Ua

using a significance level of α = 0.05. Assume df = 100.

- t= -1.89; there is insufficient evidence to conclude that the mean scores differ for mid-level and upper-level managers.
- t=-1.89; reject the H0 and conclude that the mean scores differ for mid-level and upper-level managers.
- t=-1.31; there is insufficient evidence to conclude that the mean scores differ for mid-level and upper-level managers.
- t=-1.74; reject H0 and conclude that the mean scores differ for mid-level and upper-level managers.
- t=-1.74; there is insufficient evidence to conclude that the mean scores differ for mid-level and upper-level managers.

**6. Select the most appropriate answer.**

For 12 pairs of females, the reported means are 24.8 on the well-being measure for the children of alcoholics and 29.0 for the control group. A t test statistic of 2.67 for the test comparing the means was obtained. Assuming that this is the result of a dependent-samples analysis testing for a difference between the group means, report the P-value.

- 0.01 < P-value < 0.02
- 0.005 < P-value < 0.01
- 0.0076
- 0.0152
- 0.02 < P-value <0.05

7. A test for abstract reasoning is given to a random sample of students before and after they complete a formal course in logic. Calculate the test statistic for testing that the course improves the test scores assuming that d=after-before, Xbar_{d} = -3.7 and s_{d} = 4.945, n = 10 and α = 0.05. State your conclusion in terms of the problem

- t= 0.75; fail to reject the null hypothesis and conclude that the average scores on the abstract reasoning test are the same before and after course in logic
- t= 2.37; reject the null hypothesis and conclude that the course does improve the average score on the abstract reasoning test.
- t= 2.37; fail to reject the null hypothesis and conclude that the average scores on the abstract reasoning test are the same before and after the course in logic.
- t= 0.75; fail to reject the null hypothesis. There is no evidence to conclude that the course improves the average on the abstract reasoning test.
- t=2.37; fail to reject the null hypothesis. There is no enough evidence to conclude that the course improves the average score on the abstract reasoning test.

8.**Provide an appropriate response.**

You are interested in determining whether there is a difference in the mean calorie content of a serving of fries versus a serving of onion rings at fast food restaurants. Based on a sample of seventeen french fry choices from fast food restaurants, the mean caloric content is 543.35 with a standard deviation of 112.18. A sample of eight onion ring choices from fast food restaurants has a mean caloric content of 526.25 with a standard deviation of 142.32. Assuming both populations are normal with equal standard deviations, what is the test statistic for testing whether the mean caloric content is the same for french fry orders as for onion rings at fast food restaurants?

- 0.14
- 3.46
- 0.33
- 0.30
- None of these

9.**Provide an appropriate response.**

Do motivation levels between mid-level and upper-level managers differ? A randomly selected group of each were administered a survey, which measures motivation for upward mobility. The scores are summarized below:

Upper -Level | Mid-level | |

Sample size | 73 | 109 |

Mean score | 77.4 | 79.71 |

Standard Deviation | 10.6 | 6.43 |

Assuming equal population standard deviations, find the P-value for testing that the mean scores differ for upper-level and mid-level managers. Interpret using a 5% significance level

a. P-value = 0.03; since the P-value < 0.05, we reject the null hypothesis

b. P-value = 0.08; since the P-value > 0.05. we fail to reject the null hypothesis

c. P-value =0.04; since the P-value < 0.05. we reject the null hypothesis

d. P-Value =0.06; since the P-value > 0.05, we fail to reject the null hypothesis.

10..**Construct the indicated confidence interval for the difference between the two population means. Assume that the assumptions and conditions for inference have been met.**

The table below contains information pertaining to the gasoline mileage for random samples of trucks of two different types. Find a 95% confidence interval for the difference in the means μ_{X} – μ_{Y}.

Brand X | Brand Y | |

Number of trucks | 50 | 50 |

Mean mileage | 20.1 | 24.3 |

Standard deviation | 2.3 | 1.8 |

- (-5.02, -3.38)
- (20.1,24.3)
- (3.7, 4.7)
- (-4.7, -3.7)
- (3.38, 5.02)

11.**Provide an appropriate response.**

The weights before and after 9 randomly selected participants followed a particular diet were recorded. The mean of the before weights was 170.4444, the mean of the weights following the diet was 160.5556 and the standard error of the differences was 3.1333. Calculate the appropriate test statistic for testing that the average weight was lower following the diet and state your conclusion using a significance level of 0.01.

- t=3.16; reject the null hypothesis and conclude that the diet is effective for weight loss
- t= 9.47; fail to reject the null hypothesis; there is not enough information to conclude that the diet is effective for weight loss
- t=3.16; accept the null hypothesis and conclude that there is no difference in the average weight before and after the diet
- t=3.16; fail to reject the null hypothesis; there is not enough information to conclude that the diet is effective for weight loss.
- t= 9.47; reject the null hypothesis and c