1 Find the critical value z that corresponds to a confidence

1. Find the critical value z that corresponds to a confidence level of 98%. 2. A schools surveys 40 randomly selected students and asks the number of dollars spent on music during the calendar year. The results are shown below. 20 60 15 25 65 100 150 200 100 60 5 70 160 50 100 0 75 120 80 120 80 50 12 300 75 75 100 40 40 85 80 140 100 45 100 65 10 80 70 10 Using excel construct a 95% confidence level for the mean dollars spent 3. The ages (in years) of 20 randomly sampled dogs are shown in the table below. 4 2 7 1 0.5 3 11 6 8 3 1 4.5 5 3 9 10 8 2 0.5 4 Using excel construct a 80% confidence interval using a t- distribution for the mean age. 4. In the survey of 427 employees , 68 said they carpooled at least part of the year. Using excel construct a 99% confidence level for the portion of the employees that carpool. 5. A company wants to estimate the nuber of sales calls received daily. They want to be within 10 of the true mean when using a confidence interval of 95%. The standard deviation estimated is at 20. How many daily calls must they sample? 6. A community wants to estimate the portion of residents who are pet owners. They want to be within 8% of the true portion when using a confidence level of 90%. How many residents must be sampled of no perliminary estimate is available?

Solution

1. Find the critical value z that corresponds to a confidence level of 98%.

here,
          
alpha/2 = (1 - confidence level)/2 =    0.01      

Thus, by table/technology,

z(alpha/2) = critical z for the confidence interval =    2.326347874   [ANSWER]

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2. A schools surveys 40 randomly selected students and asks the number of dollars spent on music during the calendar year. The results are shown below. 20 60 15 25 65 100 150 200 100 60 5 70 160 50 100 0 75 120 80 120 80 50 12 300 75 75 100 40 40 85 80 140 100 45 100 65 10 80 70 10 Using excel construct a 95% confidence level for the mean dollars spent

Note that              
              
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 =    0.025          
X = sample mean =    78.3          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    57.34698704          
n = sample size =    40          
              
Thus,              
              
Lower bound =    60.52831116          
Upper bound =    96.07168884          
              
Thus, the confidence interval is              
              
(   60.52831116   ,   96.07168884   ) [ANSWER]

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