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In Original Source 2, "Development and initial validation of the Hangover Symptoms Scale: Prevalence and correlates of hangover symptoms in college students," one of the questions was how many times respondents had experienced at least one hangover symptom in the past year. Table 3 of the paper (page 1,445) shows that out of all 1,216 respondents, 40% answered that it was two or fewer times. Suppose the researchers are interested in the proportion of the population who would answer that it was two or fewer times. Can they conclude that this population proportion is significantly less than half (50% or 0.5)? Go through the four steps of hypothesis testing for this situation for women only. Use appropriate software or a calculator to find the standard deviation and the test statistic. (Round your standard deviation to three decimal places and your test statistic to two decimal places.) Test Statistic?

Sagot :

Only twice or less, according to 40% of respondents. Suppose the researchers are curious about the percentage of the population that would respond that it was two or less times, in which case the test statistic would be -0.208 and the standard deviation would be 0.048.

H0: The population proportion of women who have experienced two or fewer hangover symptoms in the past year is equal to or greater than 0.5.

Ha: The population proportion of women who have experienced two or fewer hangover symptoms in the past year is less than 0.5.

Set the alpha level

α = 0.05

Calculate the test statistic

n = 618 (women only)

p = 0.4 (from Table 3)

Standard Deviation = sqrt[p*(1-p)/n]

Standard Deviation = sqrt[0.4*(1-0.4)/618]

Standard Deviation = 0.048

Test Statistic = (0.4 - 0.5) / 0.048

Test Statistic = -0.208

Step 4: Make a decision

Since the test statistic (-0.208) is less than the alpha level (0.05), we reject the null hypothesis and conclude that the population proportion of women who have experienced two or fewer hangover symptoms in the past year is less than

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