{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

STAT 2103 Class Topics Chapter 6 Section 6_6 Chapter7 Chapter 10 and 11 McGlave_Benson_Sincich ver2(

# STAT 2103 Class Topics Chapter 6 Section 6_6 Chapter7 Chapter 10 and 11 McGlave_Benson_Sincich ver2(

This preview shows pages 1–5. Sign up to view the full content.

on 1 = proportion of items in the success category. f items in the failure category. e of n 1 items, let X 1 be the number of items from the success category. = , on 2= proportion of items in the success category. 2 items, let X 2 be the number of items from the success category. = , STAT 2103 Class Topics Chapter 6, Section 6.6 Testing of Hypotheses about the Binomial Proportion p The methodology requires a formulation of the null and alternative Hypotheses. In any hypothesis testing problem, only one of the following formulations will be suitable for testing. H 0 : p p 0 against H a : p > p 0 ( one-sided) Or H 0 : p p 0 against H a : p < p 0 ( one-sided) Or H 0 : p = p 0 against H a : p p 0 ( two-sided). In the hypotheses, p 0 is a specified value for p. For either formulation, the test statistic is: , For one sided alternative, P-value = P(z-distribution > the absolute value of z-statistic). For two sided alternative, P-value = Twice the P-value for one-sided alternative. Decision rule: Reject H 0 if P-value < , level of significance. α Hippocrates magazine (May/June, 1989) reported that 30% of American children skip breakfast. Suppose a foreign social scientist believes that the rate is higher for children in her country. What can we conclude in this regard if a sample of 500 children revealed that 172 skip breakfast? Test at the .05 significance level. Using StatCrunch and following the same steps as above until you will select hypothesis test. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Click on next and enter the value for the Null: prop=.30, and select “>” for the Alternative. 2
Click on Calculate Hypothesis test results: p : proportion of successes for population H 0 : p = 0.3 H A : p > 0.3 Proportio n Coun t Tota l Sample Prop. Std. Err. Z-Stat P-value p 172 500 0.344 0.020493902 2.14698 0.0159 Since the p-value of 0.0159 is lower than alpha of .05 the social scientist can conclude that the rate of children who skip breakfast in her country is higher than 30%. Chapter 7 Confidence Interval for Estimating the Difference Between Two Means μ A (1 – )100% degree of confidence interval for estimating α μ(1)- μ(2) is: Margin of Error, ME is to the right of, - X1 X2 - ± X1 X2 ME Comparing Means of Two Populations Comparison of two populations’ means is required when seeking answers to questions of the following types: Is it true that those who eat high-fiber cereal for breakfast consume on average fewer calories at 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
lunch than those who don’t eat high-fiber cereal? The mean caloric intakes may be denoted by μ 1 for those who eat high-fiber cereal and by μ 2 for those who don’t eat high-fiber cereal. The question then can be formulated as a hypothesis testing: H a : (μ 1 - μ 2 ) < 0 against the null H 0 : (μ 1 - μ 2 ) 0. In general let, μ 1 and μ 2 denote the means of two populations. One of the following formulations will be formulated for testing. H 0 : (μ 1 - μ 2 ) ≤∆ H a : (μ 1 - μ 2 ) > ; one-sided alternative H 0 : (μ 1 - μ 2 ) ≥∆ H a : (μ 1 - μ 2 ) < ; one-sided alternative H 0 : (μ 1 - μ 2 ) = H a : (μ 1 - μ 2 ) ≠∆ ; Two-sided alternative Regardless of the formulation of hypotheses, the test statistic to be computed is: , d.f. = (n 1 + n 2 – 2), where =, For one sided alternative,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}