6. Suppose the GPA of all students in this University follows a normal distribut

Question

6. Suppose the GPA of all students in this University follows a normal distribution. From a class with 25 students, we find the class average GPA is 2.85 with standard deviation 0.36. Give a 90% confidence interval for the mean GPA of all. 7. Over the three-day period from April 1 to April 3, 2016, a national poll surveyed 1500 American households to gauge their levels of discretionary spending. The question asked was how much the respondent spent the day before; not counting the purchase of a home, motor vehicle, or normal household bills. For these sampled households, the average amount spent was x = $95 with a standard deviation of s = $185, Find a 95% confidence interval for the true daily discretionary spending.

Explanation / Answer

Ans:

6)As,population standard deviation is unknown and n<30,we will use t distribution.

sample size,n=25

df=25-1=24

t value for 90% CI and df=24 is 1.711

90% Confidence interval for true mean

=2.85+/-1.711*0.36/sqrt(25)

=2.85+/-0.12

=(2.73, 2.97)

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