*A new drug has been developed to reduce the symptoms of Shingles (inflammation and pain). A group of 30 patients with Shingles were randomly assigned to one of two treatment conditions (treatment as usual and treatment with the new drug). To measure the pain and inflammation in these patients, a composite score was obtained that provides an overall discomfort score. Higher scores indicate more discomfort. Past research has shown that the overall discomfort score is normally distributed. Using an alpha of 0.01, test whether the overall discomfort score is different between the two treatment groups. The data for the two groups are provided below. For this test, you can assume that the variances are homogeneous.*

**What is the appropriate test?**-
*Tobacco use is still a leading cause of death in the United States. Given it is a preventable cause of death, it is important to reduce the number of people who use tobacco. Researches at Health Prevention College want to test a new intervention that uses Motivational Interviewing to increase people’s readiness to reduce tobacco use. To test the new intervention, they recruit 20 smokers and ask about their readiness to stop using tobacco. These smokers are then given the new intervention. Sixty days after the intervention, they follow-up with the subjects and ask them again about their readiness to stop smoking.Readiness scores range from 0 (Not ready) to 30 (Extremely ready). It is assumed that readiness is normally distributed. Using an alpha of 0.05, test whether the intervention had an impact on readiness to stop using tobacco. The data for the two time points provided below.*

**What is the appropriate test?** *A fourth grade teacher was interested in how well her students remembered what they had Iearned over a long semester break. Se gave a 40-item quiz to 11 students at the end of the semester in December, and again at the beginning of the nest semester in January. Scores for the 11 students are recorded below. Conduct an appropriate statistical text using SPSS and α = 0.05 to determine whether memory differed before and after semester break. Submit all written steps and the actual SPSS output.(worth 4.5 points)**A teacher wishes to study the amount of time students in his statistics course spend each week in study for the course. He believes that the average should be the nominal 6 hours (two hours outside class for every hour in class). So he has the students keep track of and report the time spent in study during a typical week. A total of 9 students respond.The average time spent is 6.5hours with a standard deviation of 2 hours. Is this enough evidence to show that the average time is not 6 hours of studying ?**A hospital wants to see how many gloves nurses use over the course of their shift, so the nurses keep track of their usage of gloves. Thir results are given in the table below. Is there enough evidence to support the claim that this hospital uses significantly more gloves than the expicted average of 50 ?**Assume that the weight of cereal in a 10 ounce bos in N(μ,σ*^{2}). To test whether the mean weight is bigger than 10.1 we take random sample of size n = 16 and observe that x̄ = 10.4 and s = 0.4. Test this claim on a level of α = 0.05 significance level. Does this seem to be a reasonable assumption? (Define your critica region and test statistic just as you would before except use the student t distribution.)

**Question and Answer**

Ans:T - test

**Is it one or two-tailed?**

Ans : two-tailed test

**Write the null and alternative hypotheses, perform the statistical test and write your conclusion?**

**Null Hypo :** Average Discomfort score of Treatment as usual is equal to Treatment with drug.

**Alternative Hypothesis: **Average Discomfort score of Treatment as usual is not equal to Treatment with drug.

ybar = average of treatment as usual

xbar = average of treatment with New Drug

Assumed difference

d = 0

t = [ (Ybar - Xbar) - d ] / Sp*sqrt(1/n1+1/n2)

t = 2.55

Degrees of freedom = n1 + n2 -2 = 30 -2 = 28

Critical value at alpha =0.05 with 28 degrees of freedom is 2.048

Since the calculated t value is 2.55 which is greater than critical value 2.048 hence we reject null hypothesis and conclude that Average Discomfort score of Treatment as usual is not equal to Treatment with drug.

Ans : Two sample t-test.

**Is it one or two-tailed?**

Ans : two-tailed test.

**Write the null and alternative hypotheses, perform the statistical test and write your conclusion?**

Ans :

**Null Hypothesis :**There is no difference between Baseline Readiness and 60 day readiness.

**Alternative Hypothesis:**There is difference between Baseline Readiness and 60-day readiness.

Since the calculated t value 0.780 is less than the critical value 2.024, hence we accept null hypothesis and conclude that there is no difference between Baseline Readiness and 60 day readiness.

Since the calculated t value 0.780 is less than the critical value 2.024, hence we accept null hypothesis and conclude that there is no difference between Baseline Readiness and 60 day readiness.

**Step 1:** Aim: To check if there is a diference in memory before and after the semester break.

**Step 2:** *Null Hypothesis:* There is no difference in memory before and after the semster break.

*Alternative Hypothesis:* There is a significant difference in memory before and after the semster break

**Step 3 :** Level of significance : 0.05

**Step 4:** Data Collection :

**Step 5:** Test Statistic:

We will use paired t-test:

**Step 6 :** Critical value at 95% confidence interval or 0.05 level of significance is 2.228

**Step 7:** Acceptance and Rejection region:

Since the calculated t- value 2.211 which is less than critical region 2.228, Hence we accept the null hypothesis.

**Step 8 :** Conclusion : Since the P-value is 0.051 which is more than 0.05, hence we can conclude that their is no difference Before and after.

μ = 6

Xbar = 6.6

n = 9

Standard Deviation (SD)=9.17

Standard Error (S.E) = 3.057

Degrees of freedom(DF) =8

t = ( Xbar - μ ) / SE

t =0.196

Level of significance % =0.05

Critical Value =2.306

Conclusion: Since the calculated value is 0.196 is less than 2.306, hence we will accept null hypothesis and conclude that average time is equal to 6.

*Null hypo:* mu = 50

*Alternative Hypo:* mu not equal to 50

Average = 57.375 Total = 10099.38

SD = sqrt(10099.38 / 39) = 16.09

SE = 16.09 / sqrt(39) = 2.58

xbar = average = 57.375

t = (xbar - mu) / SE = (57.375 - 50) / 2.58

= 2.85

Critical value at n-1 or 40-1 =39 degrees of freedom at level of significance 0.05 is 2.021

Conclusion Since the calculated value is 2.85 greater than critical value 2.021, hence we reject null hypothesis and conclude that hospital uses significantly more gloves than the expected average of 50.

Given,

μ = 10.1

Xbar = 10.4

n = 16

Standard Deviation (SD) = 9.17

Standard Error (S.E) = 2.2925

Degrees of freedom(DF)

t = (Xbar - μ )/S.E

t = 0.13