Brunt, Rhee, and Zhong (2008) surveyed 557 undergraduate college students to exa

Question

Brunt, Rhee, and Zhong (2008) surveyed 557 undergraduate college students to examine their weight status, health behaviors, and diet. Using body mass index (BMI), they classified the students into four categories: underweight, healthy weight, overweight, and obese. They also measured dietary variety by counting the number of different foods each student ate from several food groups. Note that the researchers are not measuring the amount of food eaten, but rather the number of different foods eaten (variety, not quantity). Nonetheless, it was somewhat surprising that the results showed no differences among the four weight categories that were related to eating fatty and/or sugary snacks. Suppose a researcher conducting a follow up study obtains a sample of n = 25 students classified as healthy weight and a sample of n = 36 students classified as overweight. Each student completes the food variety questionnaire, and the healthy-weight group produces a mean of M = 4.01 for the fatty, sugary snack category compared to a mean of M = 4.48 for the overweight group. The results from the Brunt, Rhee, and Zhong study showed an overall mean variety score of mu = 4.22 for the discretionary sweets or fats food group. Assume that the distribution of scores is approximately normal with a standard deviation of sigma = 0.60. Does the sample of n = 36 indicate that the number of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean? Use a two-tailed test with alpha = .05.

Explanation / Answer

Given that,
population mean(u)=4.22
standard deviation, =0.6
sample mean, x =4.48
number (n)=36
null, Ho: =4.22
alternate, H1: !=4.22
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 4.48-4.22/(0.6/sqrt(36)
zo = 2.6
| zo | = 2.6
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.6 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.6 ) = 0.00932
hence value of p0.05 > 0.00932, here we reject Ho
ANSWERS
---------------
null, Ho: =4.22
alternate, H1: !=4.22
test statistic: 2.6
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00932

it it significantly diffrent from population mean

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