Question

A researcher conducts an independent groups study examining the effectiveness of diet type on a sample of overweight adults. An independent-measures t-test is used to analyze the dada. one-half of the adults receive a vegetarian diet and the other half receives a low carb, protein-rich diet. the two groups have equal sample sizes. After 6 weeks of dieting, the researcher records body weight information (in pounds) for both groups. the data are as follows ( assume that the difference between the two population means, equal zero.

df-2

vegetarian lowcarb

n=20 n=20

M=142 M=136

S²= 82.4 S²=56.2   

A) Does type of diet have a statistically significant effect? use an alpha level of .05, two-tailed  (the table of critical t values)??

B) What is the value of the critical t statistics ??

C) What is the value of the obtained t statistic ??

Answer 1

n1 = 20

= 142

s1^2 = 82.4

n2 = 20

= 136

s2^2 = 56.2

Claim: The two ponds have the same mean salinity value.

The null and alternative hypothesis is

For doing this test first we have to check the two groups have population variances are equal or not.

Null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

F = 82.4 / 56.2 = 1.47

Degrees of freedom => n1 - 1 , n2 - 1 => 20 - 1 , 20 - 1 => 19 , 19

Critical value = 2.168    ( Using f table)

Critical value > test statistic so we fail to reject null hypothesis.

Conclusion: The population variances are equal.

So we have to use here pooled variance.

B) Degrees of freedom = n1 + n2 - 2 = 20 + 20 - 2 = 38

Critical value = 2.024    ( Using t table)

| t | > critical value we reject null hypothesis.

C) Test statistic is