4. The chi-square test for goodness of fit No preference AaAa A developmental ps

Question

4. The chi-square test for goodness of fit No preference AaAa A developmental psychologist is studying bonding between healthy newborn babies and immediate family members. He wants to know if mothers use smell to recognize their one-week-old infants. To investigate, he selects a random sample of mothers of one-week-old infants. Each mother is presented with a garment worn by her infant and two garments wam by unrelated babies. He asks each of the mothers to identify her infant's garment. The resulting data are summarized in the following table that shows the number of mothers who selected each of the three garments. Actual Unrelated Unrelated Baby #21 10 Baby Baby #1 The developmental psychologist wants to know if mothers select the garment of their own one-week-old infant more often than might occur by chance. He plans to use a chi-square test for goodness of fit to test his hypothesis. The null hypothesis, that the mothers have no preference for the different garments, can be represented by the following, where the values specify the proportion of the population of mothers who would orient toward each of the garments: Actual Unrelated Unrelated Baby Baby #1 Baby #2 Fill in the following table with the expected frequencies if the null hypothesis is true Actual UnrelatedUnrelated Baby Baby #1 Baby #2 The chi-square statistic is X2 The distribution of the chi-square statistic has degree(s) of freedam. Use the following Distributions tol to find the critical value Chi-Square Distribution Degrees or Freedom = 3 10 11 with .05, the critical value is Since the chi-square statistic x2 is the critical value, the psychologist the null The psychologist conclude that some mothers use smell to recognize their one-week-old infants.

Explanation / Answer

applying chi square goodness of fit test:

from above expected frequency:

chi square test statistic =8.60

has 2 degree of freedom

the critical value =5.991

since chi square test statistic is greater than crtiical value ; the psychologist reject the null hypothesis

the psychologist have sufficient evidence to conclude.................

observed Expected Chi square category Probability O E=total*p =(O-E)^2/E actual 1/3 22.000 13.33 5.63 unelated baby#1 1/3 8.000 13.33 2.13 unelated baby#2 1/3 10.000 13.33 0.83 1 40 40 8.60

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