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PH 3200: Introduction to Biostatistics Module 9 Activity Not getting enough sleep is a common complaint from college students. Are sleep problems common for American women as well? In a discussion in a health class, one of the students in the class says that only about 50% of American women have trouble sleeping. You don’t think that this is correct. You are fairly certain that the student’s value is too low, but you decided to be cautious and make a two-sided test to see if the proportion of American women that have trouble sleeping is different from 50%. You decide to go to the General Social Survey page to research the topic. (1) Go to the GSS web site ( sda.berkeley.edu/GSS website. Under the analysis tab, type SLPPRBLM as a row variable and YEAR(2014) SEX(2) as selection filters, then run the table. Copy and paste the frequency distribution into your answer below. This variable asks respondents if they often or sometimes had trouble falling to sleep or staying asleep in the past year. Use “often” and “sometimes” responses for trouble falling/staying asleep. Is there evidence to show that the population proportion of American women who often or sometimes have trouble sleeping or staying asleep in the past year is different from 0.5? Use a significance level of α = 0.05. 1) Assumptions: Categorical data (trouble sleeping yes or no) Randomization: yes, random samples from GSS Large enough sample sizes so that there are at least 15 “successes” and at least 15 “failures” in each group . o np 0 = 638*0.5 = 319 o n(1-p 0 ) = 638* 0.5 = 319 so ^ p ~ N 2) Hypotheses: H 0 : p = 0.5 H a : p ≠ 0.5 3) Test Statistic ^ p = 384 638 = 0.6019 sample proportion Test Statistic z: z = ^ p p 0 p 0 ( 1 p 0 ) n
= 0.6019 0.5 0.5 ( 1 0.5 ) 638 = 5.15 Null value :0.5 Null SE: 0.0198 Test statistic z: 5.15 4) P-value Since it is a two sided hypothesis, find the area above 5.15 and the area below -5.15. Both of these areas are close to zero, so p < 0.001 or p-value 0 5) Conclusion We reject the null hypothesis. The sample supports the alternative hypothesis. There is strong evidence to show the population proportion of women who have trouble sometimes or often falling to asleep or staying asleep in the past year is different from 0.5 with a p value of almost zero (2) Use the “Inference for a Proportion” application to check your answers. Enter the sample size and number of successes. Choose Significant Test and enter 0.5 as the Null value. Copy and paste the image How did your calculations compare? Exactly the same as calculated values
Go to the GSS web site ( sda.berkeley.edu/GSS website and download the data for the variable AFTERLIF for 2008. This variable asks respondents if they believe in life after death with four possible responses: “Yes, definitely;” “Yes, probably;” “No, probably not;” and “No, definitely not.” Under the analysis tab, type AFTERLIF as a row variable and YEAR(2008) as a selection filter, then run the table. Copy and paste the frequency distribution into your answer below. Use these results to estimate the proportion of people in 2008 who believed in life after death (responses 1 and 2). Proportion of Successes: sample proportion ( ^ p ) who believe in life after death: 1024/1303 = 0.7859 Conduct a test to determine whether a majority of people believe in life after death. 1) Assumptions Categorical data (trouble sleep yes or no) Randomization: yes, random samples from GSS Large enough sample sizes so that there are at least 15 “successes” and at least 15 “failures” in each group . o np 0 = 1303*0.5 = 651.5 o n(1-p 0 ) = 1303* 0.5 = 651.5 2) Hypotheses H 0 : p = 0.5 H 0 : p > 0.5 3) Test Statistic n = 1303 ^ p = 1024 1303 = 0.7859 Test Statistic: z = ^ p p 0 p 0 ( 1 p 0 ) n
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= 0.7859 0.5 0.5 ( 1 0.5 ) 1303 ¿ 0.29 0.0139 = 20.64 4) P-value Find the area above 20.64. p < 0.001 or p 0 5) Conclusion We reject the null hypothesis; the sample supports the alternative hypothesis. There is evidence to support that the proportion of people who believe in life after death is > 50% Use the “Inference for a Proportion” application to check your answers. Enter the sample size and number of successes. Choose Significant Test and enter 0.5 as the Null value.