Question

Many businesses rent a limousine to chauffeur important clients to/from airport, hotel, or office. A random sample of the cost (in dollars) of renting a limousine for an entire day in Cincinnati was obtained. The sample size was 51, the sample mean was 410.25 and the sample standard deviation was 35.07. Let μ be the population mean of the cost of renting a limousine for businesses for an entire day in Cincinnati.

1. Obtain a point estimate for μ and its standard error.

2. Construct a 95% two-sided confidence interval for μ.

3. Construct a 99% one-sided lower confidence interval for μ.

4. Test the hypothesis μ = 400 against μ $\small \neq$ 400 using a significance level of α = 0.05. Provide all the details including test statistic, rejection region and the decision of whether to reject H0.

5. Test the hypothesis μ ≤ 400 against μ > 400 using a significance level of α = 0.01. Provide all the details including test statistic, rejection region and the decision of whether to reject H0.

Answer 1

Solution: Sample size n 51 Sample mean - 410.25 Sample Standard Deviation S 35.07 Degrees of freedom-df = n-1 = 51-1 = 50 Point estimate for μ is equal to the value of sample mean Hence, Point estimate for μ = 410.2 Standard error: s 35.07 Vn 51 SE=-= 4.911 95% confidence interval for μ For 95% confidence interval, α = 1-0.95 = 0.05 a/2 0.05/2 0.025 From t-distribution table, the value of t having an area of (a/2 0.025) in its upper tail and for (df - o.025 2.009 50) is given by- Margin of Error MoE to.025 35.07 51 MoE 9.864 95% confidence interval for μ Lower Limit of 95% confidence interval-410.25-9.864 Upper Limit of 95% confidence interval-410.25 + 9.864 Hence, 95% confidence interval-400.386 420.114 400.386 420.114