Thirty anorexic girls were treated with cognitive behavioral therapy (CBT). They were weighed before and after the treatment. The weight gains, which are positive scores, and weight losses, which are negative scores, are given below. A gain of more than zero pounds would confirm whether the treatment worked. Did the group gain more than zero pounds? Data: 1.7 0.7 –0.1 –0.7 –3.5 14.9 3.5 17.1 –7.6 1.6 11.7 6.1 1.1 –4.0 20.9 –9.1 2.1 –1.4 1.4 –0.3 –3.7 –0.8 2.4 12.6 1.9 3.9 0.1 15.4 –0.7 0 Report the appropriate null and research hypotheses and the critical value (alpha = 0.05). In addition, provide your statistical conclusion, including calculation of test statistic

Step 1 of the 5 step hypothesis testing process is to state the appropriate null and research hypotheses. They are:

H0:The population average difference = 0.

H1:The population average difference >0.

Next we calculate the means (or average) and standard deviation as well as n:

= 2.906667 = mean of the sample differences

s = 7.202343 = standard deviation of the differences

n = 30 = sample size of the differences

Now that you have some data collect, you can move on to step 2 of the 5-step hypothesis testing process: Decide what type of test it is, and what significance level is required. Since the sample size involved in your population is large (generally >=30 is the rule), a Z-test is appropriate. We will choose to use a 95% confidence level, our p=.05.

Based on this, our critical value is 1.96 (we obtain this from a normal distribution table).

Step 3 requires us to state the Decision Rule. The decision rule is:Reject H0 if the value of the z-score greater than 1.96.

Step 4 is for us to calculate the test statistic:

Step 5 is to state the conclusion. Since the value of Z = 1.96 our critical value calculated in step 3, we would reject the null hypothesisand conclude that there is evidence that the population average difference in weight gain is greater than zero. This indicates that the CBT results in significantly more weight gain.

You obtain the data shown below for 30 girls with anorexia who were treated with psychodynamic therapy. These values can be interpreted as described in Week 6: Assignment 2. Using these data and those from Week 6: Assignment 2, compare the outcomes for CBT and psychodynamic therapy. Note that you may use SPSS to compute any descriptive statistics you need, but you must compute the significance test by hand. Data: 0.3 1.7 –1.2 2.2 –1.8 0.9 2.6 –3.2 0 –0.5 –4.2 –8.6 0.1 0.6 8.2 0 –2.8 –5.0 0.8 5.0 5.8 0.3 0.6 –5.9 1.2 –7.8 –9.2 1.3 1.9 0.4 Report the appropriate null and research hypotheses and the critical value (alpha = .05). In addition, provide your statistical conclusion, including calculation of test statistic, in APA format and conceptual conclusion in one to two sentences. I don't understand any of this!! Any way to simplify how to do this??

We proceed in a similar way to the work from the previous section:

Step 1 of the 5 step hypothesis testing process is to state the appropriate null and research hypotheses. They are:

H0:The population average difference between CBT and PT equal 0.

H1:The population average difference between CBT and PT not equal to 0.

Next we calculate the means (or average) and variances for each treatment group:

CBT

= 2.906667 = mean of the sample differences

= 51.87375= variance of the differences

n1 = 30 = sample size of the differences

PT

= -0.5433333= mean of the sample differences

= 15.8853 = standard deviation of the differences

n2 = 30 = sample size of the differences

Now that you have some data collect, you can move on to step 2 of the 5-step hypothesis testing process: Decide what type of test it is, and what significance level is required. Since the sample size involved in your population is large (generally >=30 is the rule), a Z-test is appropriate. We will choose to use a 95% confidence level, our p=.05.

Based on this, our critical value is 1.96 (we obtain this from a normal distribution table).

Step 3 requires us to state the Decision Rule. The decision rule is:Reject H0 if the value of the absolute value of the z-score greater than 1.96.

Step 4 is for us to calculate the test statistic:

Step 5 is to state the conclusion. Since the absolute value of Z = > 1.96 our critical value calculated in step 3, we would reject the null hypothesis and conclude that there is evidence that the population average difference in weight gain between CBT and PT is different than zero.

APA Format:

A difference in means z-score analysis was conducted between the CBT group (n=30) and the PT group (n=30). A statistically significantdifference in the weight gains among the CBT group and the PT group was found, z = 2.30, p = .022.