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I took a sample of the grade point averages for students in my class. For 25 students, the standard deviation of grade points was 0.65 and the mean was 2.89. A 95% confidence interval for the average grade point average for all students in my class is:

a. (2.64, 3.14).
c. (2.53, 3.25).
b. (2.62, 3.16).

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princessesther2011

Answer:

B. (2.62, 3.16)

Step-by-step explanation:

Confidence Interval = mean + or - (t×sd)/√n

Mean = 2.89, sd = 0.65, n = 25, degree of freedom = n-1 = 25-1 = 24, t-value corresponding to 24 degrees of freedom and 95% confidence level is 2.064

Lower limit = 2.89 - (2.064×0.65)/√25 = 2.89 - 0.27 = 2.62

Upper limit = 2.89 + (2.064×0.65)/√25 = 2.89 + 0.27= 3.16

95% confidence interval is (2.62, 3.16)

Answer: a. ( 2.64, 3.14)

Therefore at 95% confidence interval (a,b) = ( 2.64, 3.14)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 2.89

Standard deviation r = 0.65

Number of samples n = 25

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

2.89+/-1.96(0.65/√25)

2.89+/-1.96(0.13)

2.89+/-0.25

= ( 2.64, 3.14)

Therefore at 95% confidence interval (a,b) = ( 2.64, 3.14)

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