**This is a two part assignment that will be submitted within one document. (read Jackson chap 8 & 10 and Torchim 293-308)**

**Part I**

Part I checks your understanding of key concepts from Jackson and Trochim & Donnelly.

Answer the following questions:

1. Jackson even-numbered Chapter exercises (pp. 220-221; 273-275)

The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60 people used the new toothpaste for 6 months. The mean number of cavities at their next checkup is 1.5. In the general population, the mean number of cavities at a 6-month checkup is 1.73 (Æ¡ = 1.12).

a. Is this a one- or two-tailed test? This will be a two tail test because the test is to show that a prediction is made that the population will differ from the general population that does not use the product, but the researcher is not sure what direction is. Despite the fact that the makes say that the sample that use their product will have less cavities.

b. What are H0 and Ha for this study? One tail it would be

HO : : µo > µ1, or µ new tooth paste > µ general population

Ha : µo < µ1 T or µ new tooth paste < µ general population

Two tailed HO & Ha:

HO: µo ≡ µ1, or µ new tooth paste ≡ µ general population

Ha: µo ≠ µ1 T or µ new tooth paste ≠ µ general population

c. Compute z obt

Æ¡-x= Æ¡/√N 1.12/√60 = 1.12/7.745 = .145

z= X-µ/Æ¡-x = 1.5-1.73/.145 = -.23/.145 = -1.586

z= X-**µ/Æ¡ x = 1.5 µ = 1.73 Æ¡=1.12 1.5-1.73/1.12 =-.205 1.12/60 1.12/7.745 Æ¡x= .145**

**z = 1.5-1.73/.144 -.23/.145 z= -1.59 (.444) area beyond = .055**

**z (N=60) = -1.59, p >.05 (one tailed) and there for Ho is accepted and Ha is rejected. **

**Zobt = -1.59**

d. What is zcv? z= X

Æ¡-x= Æ¡/√N 1.12/√60 = 1.12/7.745 = .145

z= X-µ/Æ¡-x = 1.5-1.73/.145 = -.23/.145 = -1.586

-**µ/Æ¡ x = 1.5 µ = 1.73 Æ¡=1.12 1.5-1.73/1.12 =-.205 1.12/60 1.12/7.745 Æ¡x= .145**

**z = 1.5-1.73/.144 -.23/.145 z= -1.59 **

e. Should H0 be rejected? What should the researcher conclude?

f. Determine the 95% confidence interval for the population mean, based on the sample mean.

4. Henry performed a two-tailed test for an experiment in which N = 24. He could not find his table of t critical values, but he remembered the tcv at df = 13. He decided to compare his tobt with this tcv. Is he more likely to make a Type I or a Type II error in this situation?

6. A researcher hypothesizes that individuals who listen to classical music will score differently from the general population on a test of spatial ability. On a standardized test of spatial ability,

= 58. A random sample of 14 individuals who listen to classical music is given the same test. Their scores on the test are 52, 59, 63, 65, 58, 55, 62, 63, 53, 59, 57, 61, 60, 59.

a. Is this a one- or two-tailed test?

b. What are H0 and Ha for this study?

c. Compute tobt

d. What is tcv?

e. Should H0be rejected? What should the researcher conclude?

f. Determine the 95% confidence interval for the population mean, based on the sample mean.

A researcher believes that the percentage of people who exercise in California is greater than the national exercise rate. The national rate is 20%. The researcher gathers a random sample of 120 individuals who live in California and finds that the number who exercise regularly is 31 out of 120.

a. What is X2 obt ?

b. What is df for this test?

c. What is X2 cv?

d. What conclusion should be drawn from these results?

Subject | General |

Due By (Pacific Time) | 10/24/2014 12:00 am |

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