Question

a study compared weight loss between patients on diet a and patients on diet b. patients on diet a lost a mean of 7.8 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).

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.

$t_{stat}=1.2979$

(round to four decimal places as needed.)

find the p - value.

$p - value = \square$

(round to four decimal places as needed.)

Explanation:

Step1: Determine the degrees of freedom

The formula for degrees of freedom when population variances are equal is \(df = n_1 + n_2-2\). Here \(n_1 = n_2=400\), so \(df=400 + 400-2=798\)

Step2: Calculate the p - value

Since the test statistic \(t_{STAT}=1.2979\) and this is a two - tailed test (we are testing for a difference in means), we use the t - distribution.
The p - value for a two - tailed t - test is \(p - value = 2\times(1 - P(T<|t_{STAT}|))\)
Using a t - table or statistical software (for a t - distribution with \(df = 798\approx\infty\) (since for large \(df\), the t - distribution approaches the standard normal distribution)), \(P(Z < 1.2979)\approx0.9032\) (using standard normal table \(P(Z<1.30)=0.9032\) as an approximation for large \(df\))
\(p - value=2\times(1 - 0.9032)=2\times0.0968 = 0.1936\)

Answer:

\(p - value = 0.1936\)