newspapers Name: September 19, 2015

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1 newspapers Name: September 19, 2015 newspapers reference: Peck, 1/e, ex The report Audience Insights: Communicating to Teens (aged 12 17) ( 2009) described teens attitudes about traditional media, such as TV, movies, and newspapers. In a representative sample of American teenage girls, 41% said newspapers were boring. In a representative sample of American teenage boys, 44% said newspapers were boring. Sample sizes were not given in the report. a. Suppose that the percentages reported were based on samples of 58 girls and 41 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using α =.05. b. Suppose that the percentages reported were based on samples of 2,000 girls and 2,500 boys. Is there convincing evidence that the proportion of those who think that newspapers are boring is different for teenage girls and boys? Carry out a hypothesis test using α =.05. c. Explain why the hypothesis tests in Parts (a) and (b) resulted in different conclusions. [1] [1] Peck, Roxy ( ). Statistics: Learning from Data (with JMP Printed Access Card) (Page 479). Cengage Learning. Kindle Edition. newspapers For an HT for the difference of two proportions: Define p 1, p 2, ˆp 1, n 1, ˆp 2, n 2 Calculate ˆp 1, n 1, ˆp 2, n 2 # p1.hat # n1 # p2.hat # n2 # Check that the necessary conditions are satisfied. 1

2 # check sample-size conditions for HT or CI for the difference of two proportions # p1 and p2 are unknown so use p1.hat and p2.hat conditions.two.proportions <- function(p1.hat, n1, p2.hat, n2){ category <- c("p1.hat * n1", "(1 - p1.hat) * n1", "p2.hat * n2", "(1 - p2.hat) * n2") count <- c(p1.hat * n1, (1 - p1.hat) * n1, p2.hat * n2, (1 - p2.hat) * n2) condition <- count >= 10 results <- data.frame(category, count, condition) return(results) } State H 0, H a and α Calculate the point estimate point.estimate, SE, the test statistic z, and the P-value p.value # point.estimate # p.c <- (p1.hat * n1 + p2.hat * n2) / (n1 + n2) # combined estimate # se <- sqrt(p.c * (1 - p.c) * (1 / n1 + 1 / n2)) # z # p.value # On a sketch of the normal probability distribution, locate z and shade the region whose area is p.value State the formal conclusion of the HT and explain how you reached that conclusion State the conclusion in context. Mention the level of confidence. small samples Check For an HT for the difference of two proportions: Define p 1, p 2, ˆp 1, n 1, ˆp 2, n 2 p = proportion who think newspapers are boring Calculate ˆp 1, n 1, ˆp 2, n 2 2

3 n.boys <- 41 n.girls <- 58 p.boys < # proportion of boys p.girls < # proportion of girls Rename variables. p1.hat <- p.boys p2.hat <- p.girls n1 <- n.boys n2 <- n.girls Barplot. barplot(c(p1.hat, p2.hat), las=1, col=c("palegreen", "turquoise"), names.arg=c("boys", "Girls"), ylab="proportion", main="newspapers") Newspapers 0.4 Proportion Boys Girls Check that the necessary conditions are satisfied. conditions.two.proportions(p1.hat, n1, p2.hat, n2) category count condition 1 p1.hat * n TRUE 2 (1 - p1.hat) * n TRUE 3 p2.hat * n TRUE 4 (1 - p2.hat) * n TRUE 3

4 State H 0, H a and α H 0 : p boys p girls = 0 H 0 : p boys p girls 0 alpha < Calculate the point estimate point.estimate, SE, the test statistic z, and the P-value p.value # point.estimate point.estimate <- p1.hat - p2.hat point.estimate [1] 0.03 # se p.c <- (p1.hat * n1 + p2.hat * n2) / (n1 + n2) # combined estimate se <- sqrt(p.c * (1 - p.c) * (1 / n1 + 1 / n2)) # z z <- point.estimate / se z [1] # p.value p.value <- 2 * (1 - pnorm(z)) p.value [1] HT for the difference of two proportions # alternative = "two.sided", "less", "greater" ht.two.proportions <- function(p1.hat, n1, p2.hat, n2, alternative){ p.hat.diff <- p1.hat - p2.hat p.pooled <- (p1.hat * n1 + p2.hat * n2) / (n1 + n2) se.diff <- sqrt(p.pooled * (1 - p.pooled) * (1 / n1 + 1 / n2)) z <- p.hat.diff / se.diff if (alternative=="two.sided") p.value <- 2 * (1 - pnorm(abs(z))) else if (alternative=="greater") p.value <- 1 - pnorm(z) else if (alternative=="less") p.value <- pnorm(z) return(list(p.hat.diff=p.hat.diff, se.diff=se.diff, z=z, p.value=p.value)) } 4

5 alternative <- "two.sided" ht.results <- ht.two.proportions(p1.hat, n1, p2.hat, n2, alternative) ht.results $p.hat.diff [1] 0.03 $se.diff [1] $z [1] $p.value [1] On a sketch of the normal probability distribution, locate z and shade the region whose area is p.value library(openintro) x.max <- 5.0 y.max <- 0.4 z <- round(ht.results$z, 3) mu <- 0 sigma <- 1 p.value <- round(ht.results$p.value, 3) title <- "Hypothesis Test for the Difference\nof Two Proportions (two-tailed)" draw.normal(x.max, y.max, z, mu, sigma, p.value, title) Hypothesis Test for the Difference of Two Proportions (two tailed) N(0, 1) p = z =

6 State the formal conclusion of the HT and explain how you reached that conclusion reject.h0 <- p.value <= alpha reject.h0 [1] FALSE State the conclusion in context. Mention the level of confidence. large samples For an HT for the difference of two proportions: Define p 1, p 2, ˆp 1, n 1, ˆp 2, n 2 Calculate ˆp 1, n 1, ˆp 2, n 2 # p1.hat # n1 # p2.hat # n2 # Check that the necessary conditions are satisfied. # check sample-size conditions for HT or CI for the difference of two proportions # p1 and p2 are unknown so use p1.hat and p2.hat conditions.two.proportions <- function(p1.hat, n1, p2.hat, n2){ category <- c("p1.hat * n1", "(1 - p1.hat) * n1", "p2.hat * n2", "(1 - p2.hat) * n2") count <- c(p1.hat * n1, (1 - p1.hat) * n1, p2.hat * n2, (1 - p2.hat) * n2) condition <- count >= 10 results <- data.frame(category, count, condition) return(results) } State H 0, H a and α Calculate the point estimate point.estimate, SE, the test statistic z, and the P-value p.value 6

7 # point.estimate # p.c <- (p1.hat * n1 + p2.hat * n2) / (n1 + n2) # combined estimate # se <- sqrt(p.c * (1 - p.c) * (1 / n1 + 1 / n2)) # z # p.value # On a sketch of the normal probability distribution, locate z and shade the region whose area is p.value State the formal conclusion of the HT and explain how you reached that conclusion State the conclusion in context. Mention the level of confidence. Check For an HT for the difference of two proportions: Define p 1, p 2, ˆp 1, n 1, ˆp 2, n 2 Calculate ˆp 1, n 1, ˆp 2, n 2 n.boys < n.girls < p.boys < # proportion of boys p.girls < # proportion of girls alternative <- "two.sided" alpha < Rename variables. p1.hat <- p.boys p2.hat <- p.girls n1 <- n.boys n2 <- n.girls Check that the necessary conditions are satisfied. conditions.two.proportions(p1.hat, n1, p2.hat, n2) category count condition 1 p1.hat * n TRUE 2 (1 - p1.hat) * n TRUE 3 p2.hat * n2 820 TRUE 4 (1 - p2.hat) * n TRUE 7

8 State H 0, H a and α H 0 : p boys p girls = 0 H 0 : p boys p girls 0 alpha < Calculate the point estimate point.estimate, SE, the test statistic z, and the P-value p.value # point.estimate point.estimate <- p1.hat - p2.hat point.estimate [1] 0.03 # se p.c <- (p1.hat * n1 + p2.hat * n2) / (n1 + n2) # combined estimate se <- sqrt(p.c * (1 - p.c) * (1 / n1 + 1 / n2)) # z z <- point.estimate / se z [1] # p.value p.value <- 2 * (1 - pnorm(z)) p.value [1] Check. HT for the difference of two proportions # alternative = "two.sided", "less", "greater" ht.two.proportions <- function(p1.hat, n1, p2.hat, n2, alternative){ p.hat.diff <- p1.hat - p2.hat p.pooled <- (p1.hat * n1 + p2.hat * n2) / (n1 + n2) se.diff <- sqrt(p.pooled * (1 - p.pooled) * (1 / n1 + 1 / n2)) z <- p.hat.diff / se.diff if (alternative=="two.sided") p.value <- 2 * (1 - pnorm(abs(z))) else if (alternative=="greater") p.value <- 1 - pnorm(z) else if (alternative=="less") p.value <- pnorm(z) return(list(p.hat.diff=p.hat.diff, se.diff=se.diff, z=z, p.value=p.value)) } 8

9 alternative <- "two.sided" ht.results <- ht.two.proportions(p1.hat, n1, p2.hat, n2, alternative) ht.results $p.hat.diff [1] 0.03 $se.diff [1] $z [1] $p.value [1] On a sketch of the normal probability distribution, locate z and shade the region whose area is p.value library(openintro) x.max <- 5.0 y.max <- 0.4 z <- round(ht.results$z, 3) mu <- 0 sigma <- 1 p.value <- round(ht.results$p.value, 3) title <- "Hypothesis Test for the Difference\nof Two Proportions (two-tailed)" draw.normal(x.max, y.max, z, mu, sigma, p.value, title) Hypothesis Test for the Difference of Two Proportions (two tailed) N(0, 1) p = z =

10 State the formal conclusion of the HT and explain how you reached that conclusion reject.h0 <- p.value <= alpha reject.h0 [1] TRUE State the conclusion in context. Mention the level of confidence. Conclusion. The two HTs gave very different results because the large samples resulted in a much smaller SE, so the z test statistic was larger. 10