7. Ina statistics class, a teacher is interested in seeing if merely having students sign in everyday in class wil increase attendance. Over many prior semesters of statistics teaching, it has been found that the average number of dlasses students attend is u= 24 of 30 during the semester with = 12. One semester the teacher institutes a sign-in sheet on which students sign in each day. At the end of the semester, in a classroom of n= 49 students, the students have attended an average of M= 28 classes. Test the hypothesis that the use of the sign-in sheet improved attendance. Be sure to give all the steps outlined above. Use a = .05.

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**Part II. Hypotheses to Test**

Below are a series of hypotheses to be tested. I have done one sample to show you the format in which I want you to do all problems. Follow the example in doing each problem.

**Steps to follow in doing problems:**

Please give the following steps for each problem:

a) If not given in the problem, state the null and alternative hypotheses  
b) Indicate whether you have a one- or a two-tailed test situation  
c) Give the formula for the test statistic you will use  
d) State the null sampling distribution  
e) State a decision rule and diagram values for which you will or will not reject H₀  
f) Calculate the standard error and the test statistic, state the statistical decision to reject or not reject H₀  
g) Briefly state in words what the outcome of the test is  

---

**Example Problem:**  
You have devised a new approach to testing IQ in young children. You believe that your testing method will produce different results from the usual tests now in use. The usual tests give μ = 100 and σ = 15 in the population. You test n = 25 children with your method and get M = 113. Does your method change the IQ measure from that in the population? Use α = .05.

a) Hypotheses:  
   H₀: μ = 100  
   H₁: μ ≠ 100  

b) Two-tailed test because there is no prediction of direction

c) Formula for test statistic:  
   \( z_{\text{calc}} = \frac{M - \mu}{\sigma_{\overline{X}}} \)

d) Null sampling distribution  

e) Decision rule:  
   Reject H₀ if \( z_{\text{crit}} \) is ≤ -1.96 or ≥ 1.96

f) Standard error:  
   \( \sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{25}} = 3 \)

   Test Statistic:  
   \( z_{\text{calc}} = \frac{M - \mu}{\sigma_{\overline{X}}} = \frac{113 - 100}{3} = 4.33 \)

   Decision: Reject H₀  

g) Interpretation: Since the null hypothesis has been rejected
Transcribed Image Text:**Part II. Hypotheses to Test** Below are a series of hypotheses to be tested. I have done one sample to show you the format in which I want you to do all problems. Follow the example in doing each problem. **Steps to follow in doing problems:** Please give the following steps for each problem: a) If not given in the problem, state the null and alternative hypotheses b) Indicate whether you have a one- or a two-tailed test situation c) Give the formula for the test statistic you will use d) State the null sampling distribution e) State a decision rule and diagram values for which you will or will not reject H₀ f) Calculate the standard error and the test statistic, state the statistical decision to reject or not reject H₀ g) Briefly state in words what the outcome of the test is --- **Example Problem:** You have devised a new approach to testing IQ in young children. You believe that your testing method will produce different results from the usual tests now in use. The usual tests give μ = 100 and σ = 15 in the population. You test n = 25 children with your method and get M = 113. Does your method change the IQ measure from that in the population? Use α = .05. a) Hypotheses: H₀: μ = 100 H₁: μ ≠ 100 b) Two-tailed test because there is no prediction of direction c) Formula for test statistic: \( z_{\text{calc}} = \frac{M - \mu}{\sigma_{\overline{X}}} \) d) Null sampling distribution e) Decision rule: Reject H₀ if \( z_{\text{crit}} \) is ≤ -1.96 or ≥ 1.96 f) Standard error: \( \sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{25}} = 3 \) Test Statistic: \( z_{\text{calc}} = \frac{M - \mu}{\sigma_{\overline{X}}} = \frac{113 - 100}{3} = 4.33 \) Decision: Reject H₀ g) Interpretation: Since the null hypothesis has been rejected
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