Question

According to the website www.collegedrinkingprevention.gov, "About 25 percent of college students report academic consequences of their drinking including missing class, falling behind, doing poorly on exams or papers, and receiving lower grades overall." A statistics student is curious about drinking habits of students at his college. He wants to estimate the mean number of alcoholic drinks consumed each week by students at his college. 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. Construct the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college.

Answer 1

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, $\bar x$=3.90 σ=3.51

Now find degree of freedom.

$df=n-1\\df=50-1=49$

The confidence level is 90% and df=49

Therefore,

$\frac{\alpha}{2} =\frac{1-0.90}{2}$

$\frac{\alpha}{2} =0.05$

Now by using t distribution table look at 49 df and  alpha level on 0.05.

$t_{\frac{\alpha}{2}} = 1.67653$

Calculate SE as shown:

$SE=\frac{\sigma}{\sqrt{50} }$

$SE=\frac{3.51}{\sqrt{50}}=0.4964$

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.

$\bar x-E<\mu<\bar x +E\\3.90-0.832<\mu<3.90+0.832\\3.068<\mu<4.732$

Hence, the required confidence interval is (3.068,4.732)

Answer 2

Using the t-distribution, it is found that the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college is (3.07, 4.73).

We have the standard deviation for the sample, which is why the t-distribution is used to solve this question.

To build the interval, these following information about the sample are used:

• Mean of 3.9, thus $\overline{x} = 3.9$
• Standard deviation of 3.51, thus $s = 3.51$
• Sample size of 50, thus $n = 50$.

The first step is finding the number of degrees of freedom, which is one less than the sample size, thus df = 49.

Then, we find the critical value for a 90% confidence interval with 49 df, which using a calculator or the t-table is t = 1.6766.

The margin of error is of:

$M = t\frac{s}{\sqrt{n}}$

Then

$M = 1.6766\frac{3.51}{\sqrt{50}} = 0.83$

The confidence interval is:

$\overline{x} \pm M$

Then

$\overline{x} - M = 3.90 - 0.83 = 3.07$

$\overline{x} + M = 3.90 + 0.83 = 4.73$

Thus, the 90% confidence interval to estimate the average number of alcoholic drinks consumed each week by students at this college is (3.07, 4.73).

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