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 μ = 4.22 for the discretionary sweets or fats food group. Assume that the distribution of scores is approximately normal with a standard deviation of σ = 0.60.
a. Does the sample of n = 36 indicate that number of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean? Use a two-tailed test with α = .05.
b. The null hypothesis is H0 : ________.

Respuesta :

Answer:

Step-by-step exhplanation:

a) n = 36 , M = 4.48 ,  = 4.22 ,  = 0.60 ,  = 0.05

The hypothesis are given by,

H0 :  = 4.22 v/s H1 :  4.22

The Z statistic to test the hypothesis  is given by,

 Z = (M - μ) / (σ√n) ~ N(0,1)

Where the critical region /Z/ ≥ Zcritical for a two tailed test.

Z = (4.48 - 4.22)/(0.60√36)

Z = 2.6

Critical value = Z/2 =  1.96

The calculated value z > critical value z .

Thus, we reject the null hypothesis.

The number of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean.

b) n =25, M = 4.01 ,  = 4.22 ,  = 0.60 , = 0.05

The hypothesis is given by,

H0 :  4.22 v/s H1 :  < 4.22

The Z statistic to test the hypothesis  is given by,

Z = (M - μ) / (σ√n) ~ N(0,1)

Where the critical region /Z/ <  Zcritical for a one tailed test.

Z = (4.01- 4.22)/(0.60√25)

Z = -1.75

The critical value of z = -1.645

The calculated value z > The critical value of z

Hence we reject the null hypothesis.

The healthy-weight students eat significantly fewer fatty, sugary snacks than the overall population.

The probability shows that since z = 2.6 which is more than the critical value of 1.96, the null hypothesis will be rejected.

How to calculate probability?

From the information given, the standard error will be:

= 0.60/✓36

= 0.60/6

= 0.10

The critical region will be:

= (4.48 - 4.22) / (0.60 × ✓36)

= 2.6

The critical value at 0.05 is 1.96.

Since z = 2.6 which is more than the critical value of 1.96, the null hypothesis will be rejected.

The students consume significantly different number of fatty, sugary drinks and snacks than the overall population average.

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