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Statistical Analysis and Hypothesis Testing in Various Research Scenarios

Solved assignment covering hypothesis testing and statistical analysis in diverse research cases.

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    Statistics 5 1. A researcher is comparing the reaction times of 10 year olds who play video games and 10 - year - olds who do not. Random samples from each group provide the results shown below. Reaction times of 10 - year - olds Players of Video Games Non - Players Of Video Games Sample size 15 10 Mean reaction time (seconds) 0.50 0.52 Standard deviation of reaction times (seconds) 0.01 0.02 Assume that the reaction times for bot h populations are normally distributed with the same population standard deviation. a. Construct a 98% confidence interval for the difference between the population means of the reaction times. Note : Express your answer to 5 decimal places of accuracy. 𝑠 𝑝 = ΰΆ§ ( 𝑛 1 βˆ’ 1 ) 𝑠 1 2 + ( 𝑛 2 βˆ’ 1 ) 𝑠 2 2 𝑛 1 + 𝑛 2 βˆ’ 2 = ΰΆ§ ( 15 βˆ’ 1 ) ( 0 . 01 ) 2 + ( 10 βˆ’ 1 ) ( 0 . 02 ) 2 15 + 10 βˆ’ 2 = 0 . 01474 . The 98% confidence interval for the difference between the population means of the reaction times

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    α‰Ž ( 𝒙 ΰ΄₯ 𝟏 βˆ’ 𝒙 ΰ΄₯ 𝟐 ) Β± 𝒕 𝜢 𝟐 ⁄ , 𝒏 𝟏 + 𝒏 𝟐 βˆ’ 𝟐 𝒔 𝒑 ΰΆ¨ 𝟏 𝒏 𝟏 + 𝟏 𝒏 𝟐 ቏ = α‰Ž ( 𝟎 . πŸ“πŸŽ βˆ’ 𝟎 . πŸ“πŸ ) Β± 𝒕 𝟎 . 𝟎𝟏 , πŸπŸ‘ ( 𝟎 . πŸŽπŸπŸ’πŸ•πŸ’ ) ΰΆ¨ 𝟏 πŸπŸ“ + 𝟏 𝟏𝟎 ቏ = [ βˆ’ 𝟎 . 𝟎𝟐𝟎𝟎𝟎 Β± ( 𝟐 . πŸ’πŸ—πŸ—πŸ–πŸ• ) ( 𝟎 . πŸŽπŸŽπŸ”πŸŽπŸ ) ] = [ βˆ’ 𝟎 . 𝟎𝟐𝟎𝟎𝟎 Β± 𝟎 . πŸŽπŸπŸ“πŸŽπŸ’ ] = [ βˆ’ 𝟎 . πŸŽπŸ‘πŸ“πŸŽπŸ’ , βˆ’ 𝟎 . πŸŽπŸŽπŸ’πŸ—πŸ” ] b. Using a 2.5% significance level, can the researcher conclude that the mean r eaction time of players of video games is less than the mean reaction time of non - players of video games? Formulate and test the appropriate hypotheses. Use the critical value approach. Null hypothesis, H 0 : ΞΌ 1 β‰₯ ΞΌ 2 Alternative hypothesis, H a : ΞΌ 1 < ΞΌ 2 Level of significance, Ξ± = 0.025 The critical value for this left - tailed t - test with 𝒏 𝟏 + 𝒏 𝟐 βˆ’ 𝟐 = πŸπŸ‘ degrees of freedom at 2.5% significance level is – 2.0687. Test statistic, 𝒕 = ( 𝒙 ΰ΄₯ 𝟏 βˆ’ 𝒙 ΰ΄₯ 𝟐 ) 𝒔 𝒑 ΰΆ§ 𝟏 𝒏 𝟏 + 𝟏 𝒏 𝟐 = 𝟎 . πŸ“πŸŽ βˆ’ 𝟎 . πŸ“πŸ ( 𝟎 . πŸŽπŸπŸ’πŸ•πŸ’ ) ΰΆ§ 𝟏 πŸπŸ“ + 𝟏 𝟏𝟎 = βˆ’ 𝟎 . 𝟎𝟐 𝟎 . πŸŽπŸŽπŸ”πŸŽπŸ = βˆ’ πŸ‘ . πŸ‘πŸπŸ‘πŸ” Since – 3.3236 < – 2.0687, reject the null hypothesis. Therefore, at 2.5% significance level, the researcher can conclude that the mean reaction time of players of video games is less than the mean reaction time of non - players of video games. 2. Is there any difference between the proportion of Americans who wish they were rich and the proportion of Canadians who wish they were rich? A random sample of 1,000 Americans showed that 550 wish they were rich, while a random sample of 500 Canadians showed that 225 wished they were rich. a. Find a 97.5% confidence interval for the difference between the population

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    proportions who wish they were rich. Note : Express your answer to 4 decimal places of accuracy. 𝑝 Μ‚ 1 = 𝑋 1 𝑛 1 = 550 1 , 000 = 0 . 55 𝑝 Μ‚ 2 = 𝑋 2 𝑛 2 = 225 500 = 0 . 45 The 97.5% confidence interval for the difference between the Population proportions who wish they were rich: α‰Ž ( 𝒑 ෝ 𝟏 βˆ’ 𝒑 ෝ 𝟐 ) Β± 𝒁 𝜢 𝟐 ⁄ ΰΆ¨ 𝒑 ෝ 𝟏 ( 𝟏 βˆ’ 𝒑 ෝ 𝟏 ) 𝒏 𝟏 + 𝒑 ෝ 𝟐 ( 𝟏 βˆ’ 𝒑 ෝ 𝟐 ) 𝒏 𝟐 ቏ = α‰Ž ( 𝟎 . πŸ“πŸ“ βˆ’ 𝟎 . πŸ’πŸ“ ) Β± 𝒁 𝟎 . πŸŽπŸπŸπŸ“ ΰΆ¨ 𝟎 . πŸ“πŸ“ ( 𝟏 βˆ’ 𝟎 . πŸ“πŸ“ ) 𝟏 , 𝟎𝟎𝟎 + 𝟎 . πŸ’πŸ“ ( 𝟏 βˆ’ 𝟎 . πŸ’πŸ“ ) πŸ“πŸŽπŸŽ ቏ = [ 𝟎 . 𝟏𝟎𝟎𝟎 Β± ( 𝟐 . πŸπŸ’ ) ( 𝟎 . πŸŽπŸπŸ•πŸ ) ] = [ 𝟎 . 𝟏𝟎𝟎𝟎 Β± 𝟎 . πŸŽπŸ”πŸπŸŽ ] = [ 𝟎 . πŸŽπŸ‘πŸ—πŸŽ , 𝟎 . πŸπŸ”πŸπŸŽ ] b. Test at the 1% level of significance, whether the two population proportions are significantly different. Formulate and test the appropriate hypotheses. Use the critical value approach. Null hypothesis, H 0 : p 1 = p 2 Alternative hypothesis, H a : p 1 β‰  p 2 Level of significance, Ξ± = 0.0 1 The critical value s for this two - tailed z - test at 1 % significance level are – 2. 576 and 2.576 . Test statistic, 𝒛 = ΰ΅« 𝑝 ෝ 1 βˆ’ 𝑝 ෝ 2 ΰ΅― ΰΆ¨ 𝒑 ෝ 𝟏 ΰ΅« 𝟏 βˆ’ 𝒑 ෝ 𝟏 ΰ΅― 𝒏 𝟏 + 𝒑 ෝ 𝟐 ΰ΅« 𝟏 βˆ’ 𝒑 ෝ 𝟐 ΰ΅― 𝒏 𝟐 = 𝟎 . πŸ“πŸ“ βˆ’ 𝟎 . πŸ’πŸ“ ΰΆ§ 𝟎 . πŸ“πŸ“ ( 𝟏 βˆ’ 𝟎 . πŸ“πŸ“ ) 𝟏 , 𝟎𝟎𝟎 + 𝟎 . πŸ’πŸ“ ( 𝟏 βˆ’ 𝟎 . πŸ’πŸ“ ) πŸ“πŸŽπŸŽ = 𝟎 . 𝟏𝟎 𝟎 . πŸŽπŸπŸ•πŸ = πŸ‘ . πŸ”πŸ–

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