Suppose a researcher conducting a follow up study obtained a sample of n =25 stu

Question

Suppose a researcher conducting a follow up study obtained 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 mean =4.22 for the discretionary sweets or fats food group. Assume that the distribution of scores is approximately normal with a standard deviation of .60. A. 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. B. Based on the sample of n =25 healthy weight students, can you conclude that healthy weight students eat significantly fewer fatty sugary snacks than the overall population? Use a one tailed test with alpha= .05.

Explanation / Answer

Solution:-

Students with healthy weight :-

M = 4.01

S.D = 0.60

Overweight group:-

M = 4.48

S.D = 0.60

Test for sample of n = 36 indicate that the number of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 4.22

Alternative hypothesis: 4.22

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.60 / 6

S.E = 0.10

DF = n - 1

D.F = 35

t = (x - ) / SE

x - = 4.48 - 4.22 = 0.26

t = + 2.6

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 35 degrees of freedom is less than - 2.60 or greater than 2.60.

Thus, the P-value = 0.00678 + 0.00678 = 0.013562

Interpret results. Since the P-value (0.013562) is less than the significance level (0.05), we have to reject the null hypothesis.

From this we can conclude that we have sufficient evidence that the number of fatty, sugary snacks eaten by overweight students is significantly different from the overall population mean.

Test for healthy weight students eat significantly fewer fatty sugary snacks than the overall population.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: > 4.22

Alternative hypothesis: < 4.22

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.6 / 5

S.E = 0.12

DF = n - 1

D.F = 24

t = (x - ) / SE

t = - 1.75

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Thus the P-value in this analysis is 0.046448.

Interpret results. Since the P-value (0.046448) is less than the significance level (0.05), we have to reject the null hypothesis.

From this we can conclude that healthy weight students eat significantly fewer fatty sugary snacks than the overall population.

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