A student claims that first year college students must study

A student claims that first year college students must study 150 minutes per night during the school week. A survey finds that the average study time claimed by 101 students 142 minutes. The standard deviation from the survey is 65 minutes. Is there reason to believe that the average study time per week is less than 150 minutes based on the survey Use a 5% level of significance. Find the 90% confidence interval for average study time of students. In the absence of special preparation. SAT math scores in recent years have varied normally with a mean of 518 and a standard deviation of 114. One hundred students go through a rigorous training program designed to raise their SAT math scores. Their average SAT math score after this training was 533.7. Is there significance at the 10% level that the mean SAT score after all the students go through a rigorous training is greater than 51S? Use standard deviation of 114 as the population standard deviation.

Solution

4.

A)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   150  
Ha:    u   <   150  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    142          
uo = hypothesized mean =    150          
n = sample size =    101          
s = standard deviation =    65          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.236907769          
              
Also, the p value is              
              
p =    0.108060663          
              
As |z| < 1.649, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.  

Ther is no significant evidence that the mean time in studying is less than 150 minutes.

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b)      

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    142          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    65          
n = sample size =    101          
              
Thus,              
Margin of Error E =    10.63848845          
Lower bound =    131.3615115          
Upper bound =    152.6384885          
              
Thus, the confidence interval is              
              
(   131.3615115   ,   152.6384885   ) [ANSWER]

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