Answer:
The required confidence interval is (3.068,4.732)
Step-by-step explanation:
Consider the provided information.
He plans to use a 90% confidence interval. He surveys a random sample of 50 students. The sample mean is 3.90 alcoholic drinks per week. The sample standard deviation is 3.51 drinks and wants to construct 90% confidence interval.
Thus, n=50, =3.90 σ=3.51
Now find degree of freedom.
The confidence level is 90% and df=49
Therefore,
Now by using t distribution table look at 49 df and alpha level on 0.05.
Calculate SE as shown:
Now multiply 1.67653 with 0.4964
Therefore, the marginal error is: 1.67653 × 0.4964≈ 0.832
Now add and subtract this value in given mean to find the confidence interval.
Hence, the required confidence interval is (3.068,4.732)
