Question

1. Twenty male athletes were given a special candy bar daily for several weeks, and each man’s body weight was recorded. It is found that the mean change in body weight was +1.5 kg. The variance of these 20 data of weight-change has been estimated as s2 = 6.25 kg2. (50 pts) – t distribution and chi-square distribution tables are included.

1. (a) Test the null hypothesis that the candy bar has NO significant effect on body weight.

2. (b) Test the null hypothesis that the candy bar will have positive impact on the body weight

   change.

(c) Test the null hypothesis that H0: ?2 = 3.00 kg2
(d) Test the null hypothesis that H0: σ2 ≥ 5.00 kg2

Answer 1

a)

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

Null hypothesis: ud = 0

Alternative hypothesis: ud ≠ 0

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

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

s2 = 6.25

SE = s / sqrt(n)

S.E = 1.3975

DF = n - 1 = 20 -1

D.F = 19

t = [ (x1 - x2) - D ] / SE

t = 1.073

where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 19 degrees of freedom is more extreme than 1.073; that is, less than - 1.073 or greater than 1.073.

P-value = P(t < - 1.073) + P(t > 1.073)

P-value = 0.1485 + 0.1485

P-value = 0.297

Interpret results. Since the P-value (0.297) is greater than the significance level (0.05), we failed to reject the null hypothesis.

Do not Reject H0. Hence candy bar has NO significant effect on body weight.

b)

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

Null hypothesis: ud = 0

Alternative hypothesis: ud > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

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

s2 = 6.25

SE = s / sqrt(n)

S.E = 1.3975

DF = n - 1 = 20 -1

D.F = 19

t = [ (x1 - x2) - D ] / SE

t = 1.073

where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a one-tailed test, the P-value is the probability that a t statistic having 19 degrees of freedom is greater than 1.073

P-value = P(t > 1.073)

P-value = 0.1485

Interpret results. Since the P-value (0.1485) is greater than the significance level (0.05), we failed to reject the null hypothesis.

Do not Reject H0. Hence candy bar has no positive impact on the body weight change.

c) 95% confidence interval for the population mean μ is  C.I = (-1.425, 4.425).

C.I = 1.5 + 2.093*1.3975

C.I = 1.5 + 2.925

C.I = (-1.425, 4.425)

d)

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

Null hypothesis H0: σ2 = 3.00

Alternative hypothesis HA: σ2 ≠ 3.00

Formulate an analysis plan. For this analysis, the significance level is 0.05.

Analyze sample data. Using sample data, the degrees of freedom (DF), and the test statistic (X2).

DF = n - 1 = 20 -1

D.F = 19

We use the Chi-Square Distribution Calculator to find P(Χ2 > 39.583) = 0.004

Interpret results. Since the P-value (0.004) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that population variance is  3.00 kg2.