Question

a study compared weight loss between patients on diet a and patients on diet b. patients on diet a lost a mean of 7.6 pounds in six months, whereas patients on diet b lost a mean of 7.3 pounds in six months. suppose that the study was based on a sample of 400 patients on diet a and 400 patients on diet b, and the standard deviation of the amount lost was 3.6 pounds for diet a and 2.9 pounds for diet b. complete parts (a) through (d).
c. in the context of this study, what is the meaning of a type ii error?
a type ii error is committed when one concludes that there is a significant difference in mean weight loss between the two diets when there is not a significant difference.
a type ii error is committed when one does not reject both the null and the alternative hypothesis.
a type ii error is committed when one concludes that there is not a significant difference in mean weight loss between the two diets when there is indeed a significant difference.
a type ii error is committed when one concludes that there is not a significant difference in mean weight loss of only diet a.
d. assume the population variances are equal. using a 0.1 level of significance, is there evidence of a difference in the mean weight loss of patients between the two diets?
find the test statistic.
( t_{stat}=square )
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the pooled variance

The formula for pooled variance \(S_{p}^{2}\) is \(S_{p}^{2}=\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}\)
Here, \(n_{1} = 400\), \(S_{1}=3.6\), \(n_{2}=400\), \(S_{2}=2.9\)
\(S_{p}^{2}=\frac{(400 - 1)\times3.6^{2}+(400 - 1)\times2.9^{2}}{400+400 - 2}\)
\(=\frac{399\times12.96+399\times8.41}{798}\)
\(=\frac{399\times(12.96 + 8.41)}{798}\)
\(=\frac{399\times21.37}{798}\)
\(S_{p}^{2}=10.685\)

Step2: Calculate the test statistic

The formula for the \(t\) - test statistic for two - sample \(t\) - test (equal variances) is \(t_{STAT}=\frac{\bar{X}_{1}-\bar{X}_{2}}{S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\)
\(\bar{X}_{1} = 7.6\), \(\bar{X}_{2}=7.3\), \(n_{1}=n_{2}=400\), \(S_{p}=\sqrt{10.685}\approx3.27\)
\(t_{STAT}=\frac{7.6 - 7.3}{3.27\sqrt{\frac{1}{400}+\frac{1}{400}}}\)
\(=\frac{0.3}{3.27\sqrt{\frac{2}{400}}}\)
\(=\frac{0.3}{3.27\times\frac{\sqrt{2}}{20}}\)
\(=\frac{0.3\times20}{3.27\times1.414}\)
\(=\frac{6}{4.624}\)
\(t_{STAT}\approx1.2976\)

Answer:

\(t_{STAT}\approx1.2976\)