Thank you 6.21 Infant deaths in King County, Washington were grouped by season o
ID: 2926014 • Letter: T
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Thank you
6.21 Infant deaths in King County, Washington were grouped by season of the year. The number of deaths by season, for selected causes of death, are listed in Table 6.13 Table 6.13 Death Data for Problem 6.21 Winter Spring Summer 48 40 93 19 71 Autumn Asphyxia Immaturity Congenital malformations Infection Sudden infant death syndrome 46 50 30 95 40 78 34 35 83 43 86 40 87 (a) (b) (c) At the 5% significance level, test the hypothesis that SIDS deaths are uniformly (p 1 /4) spread among the seasons. At the 10% significance level, test the hypothesis that the deaths due to infection are uniformly spread among the seasons. What can you say about the p-value for testing that asphyxia deaths are spread uniformly among seasons? Immaturity deaths?Explanation / Answer
(a)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 4 - 1 = 3
Total SIDS deaths = 78 + 71 + 87 + 86 = 322
(Ei) = n * pi
(Ei) = 322 * 0.25 = 80.5 for i = 1,2,3,4
2 = [ (Oi - Ei)2 / Ei ]
2 = [ (78 - 80.5)2 / 80.5 ] + [ (71 - 80.5)2 / 80.5 ] + [ (87 - 80.5)2 / 80.5 ] + [ (86 - 80.5)2 / 80.5 ]
= 2.1
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and 2 is the chi-square test statistic.
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 2.1.
We use the Chi-Square Distribution Calculator to find P(2 > 2.1) = 0.5519
Interpret results. Since the P-value (0.5519) is greater than the significance level (0.05), we accept the null hypothesis and conclude that SIDS deaths are uniformaly spread among the seasons.
(b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Formulate an analysis plan. For this analysis, the significance level is 0.1. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.
Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.
DF = k - 1 = 4 - 1 = 3
Total Infection deaths = 40 + 19 + 40 + 43 = 142
(Ei) = n * pi
(Ei) = 142 * 0.25 = 35.5 for i = 1,2,3,4
2 = [ (Oi - Ei)2 / Ei ]
2 = [ (40 - 35.5)2 / 35.5 ] + [ (19 - 35.5)2 / 35.5 ] + [ (40 - 35.5)2 / 35.5 ] + [ (43 - 35.5)2 / 35.5 ]
= 10.39
where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and 2 is the chi-square test statistic.
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 10.39.
We use the Chi-Square Distribution Calculator to find P(2 > 10.39) = 0.015
Interpret results. Since the P-value (0.015) is less than the significance level (0.1), we reject the null hypothesis and conclude that Infection deaths are not uniformaly spread among the seasons and at least one of the seasons have different number of infection deaths as compared with other seasons .
(c)
Total asphyxia deaths = 50 + 48 + 46 + 34 = 178
(Ei) = n * pi
(Ei) = 178 * 0.25 = 44.5 for i = 1,2,3,4
2 = [ (Oi - Ei)2 / Ei ]
2 = [ (50 - 44.5)2 / 44.5 ] + [ (48 - 44.5)2 / 44.5 ] + [ (46 - 44.5)2 / 44.5 ] + [ (34 - 44.5)2 / 44.5 ]
= 3.48
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 3.48.
We use the Chi-Square Distribution Calculator to find P(2 > 3.48) = 0.3234
Since the P-value (0.3234) is greater than the significance level (0.05), we accept the null hypothesis and conclude that asphyxia deaths are uniformaly spread among the seasons.
Total immaturity deaths = 30 + 40 + 36 + 35 = 141
(Ei) = n * pi
(Ei) = 141 * 0.25 = 35.25 for i = 1,2,3,4
2 = [ (Oi - Ei)2 / Ei ]
2 = [ (30 - 35.25)2 / 35.25 ] + [ (40 - 35.25)2 / 35.25 ] + [ (36 - 35.25)2 / 35.25 ] + [ (35 - 35.25)2 / 35.25 ]
= 1.44
The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 1.44.
We use the Chi-Square Distribution Calculator to find P(2 > 1.44) = 0.6962
Since the P-value (0.6962) is greater than the significance level (0.05), we accept the null hypothesis and conclude that immaturity deaths are uniformaly spread among the seasons
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