The following data represent the age in weeks at which babie

The following data represent the age (in weeks) at which babies first crawl based on a survey of 12 mothers.

52 30 44 35

47 37 56 26

47 37 28 37

Mean = 39.67

St Dev = 9.56

a) Because the sample size is small, we must verify that the data come from a population that is normally distributed and the the sample size does not contain any outliers. Are the conditions for constructing a confidence interval about the mean satisfied?

No, the population is not normall distributed

Yes, the population is normally distributed and the sample does not contain any outliers.

No, the sample contains an outlier.

b) Construct a 95% confidence interval for the mean age at which a baby first crawls. Select the correct choice and fill in any answers boxes in your choice.

A) (.....,....)

B) A 95% confidence interval can not be conducted.

c) What could be done to increase the accuracy of the interval without changing the level of confidence?

Nothing can be done

Increase the sample size

Either increase or decrease the sample size

Solution

a) Because the sample size is small, we must verify that the data come from a population that is normally distributed and the the sample size does not contain any outliers. Are the conditions for constructing a confidence interval about the mean satisfied?

Yes, the population is normally distributed and the sample does not contain any outliers. [answer]

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b)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    39.66666667          
t(alpha/2) = critical t for the confidence interval =    2.20098516          
s = sample standard deviation =    9.556847458          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
              
Lower bound =    33.59453549          
Upper bound =    45.73879785          
              
Thus, the confidence interval is              
              
(   33.59453549   ,   45.73879785   ) [ANSWER]

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c)

What could be done to increase the accuracy of the interval without changing the level of confidence?

Increase the sample size [ANSWER]