Please provide me with answers for questions A and B problem #2 A. \"Carry out T

Question

Please provide me with answers for questions A and B problem #2 A. "Carry out The Seven-Steps of Hypothesis Testing Procedures For Each Case" B. Construct Proper Confidence Interval of 95% For Each Case" 1. A sample of 9 high school seniors in a school system reports a mean of 5 hours worked at part-time jobs during a recent week: The sample standard deviation was 3 hours. Do these data provide sufficient evidence to indicate that the mean for the population is less than 8 hours? Assume a normally distributed population. 2. The owner of the Ellenborough Shopping Center claims that more than 50% of the households within a three-mile radius of the shopping center have at least one member who shops at the center at least once a week. In a sample of 300 households in the area, an investigator found that members of 171 households did so. Do these data provide sufficient evidence to support the shopping center owner's claim? Let 0.01.

Explanation / Answer

(1)

(A)

Data:    

n = 9   

= 8   

s = 3   

x-bar = 5   

Hypotheses:    

Ho: ³ 8   

Ha: < 8   

Decision Rule:    

= 0.05   

Degrees of freedom = 9 - 1 = 8

Critical t- score = -1.859548033   

Reject Ho if t < -1.859548033   

Test Statistic:    

SE = s/n = 3/9 = 1

t = (x-bar - )/SE = (5 - 8)/1 = -3

p- value = 0.008535841   

Decision (in terms of the hypotheses):    

Since -3 < -1.859548033 we reject Ho and accept Ha

Conclusion (in terms of the problem):

There is sufficient evidence that < 8

(B)

n = 9     

x-bar = 5     

s = 3     

% = 95     

Standard Error, SE = s/n =    3/9 = 1

Degrees of freedom = n - 1 =   9 -1 = 8   

t- score = 2.306004133     

Width of the confidence interval = t * SE =     2.30600413329912 * 1 = 2.306004133

Lower Limit of the confidence interval = x-bar - width =      5 - 2.30600413329912 = 2.693995867

Upper Limit of the confidence interval = x-bar + width =      5 + 2.30600413329912 = 7.306004133

The 95% confidence interval is [2.694, 7.306].

(2)

(A)

Data:    

n = 300   

p = 0.5   

p' = 171/300 = 0.57   

Hypotheses:   

Ho: p 0.5   

Ha: p > 0.5   

Decision Rule:   

= 0.05   

Critical z- score = 1.644853627

Reject Ho if z > 1.644853627

Test Statistic:   

SE = {(p (1 - p)/n} = (0.5 * (1 - 0.5)/300) = 0.028867513

z = (p' - p)/SE = (0.57 - 0.5)/0.0288675134594813 = 2.424871131

p- value = 0.0076569   

Decision (in terms of the hypotheses):

Since 2.4248711 > 1.644853627 we reject Ho

Conclusion (in terms of the problem):

There is sufficient evidence that p > 0.5

(B)

n = 300    

p = 171/300 = 0.57    

% = 95    

Standard Error, SE = {p(1 - p)/n} =    (0.57(1 - 0.57))/300 = 0.028583212

z- score = 1.959963985    

Width of the confidence interval = z * SE =     1.95996398454005 * 0.0285832118559129 = 0.05602207

Lower Limit of the confidence interval = P - width =     0.57 - 0.0560220658000676 = 0.51397793

Upper Limit of the confidence interval = P + width =     0.57 + 0.0560220658000676 = 0.62602207

The 95% confidence interval is [0.514, 0.626].

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