Question

2. Results from a sample of 60 customers are as follows. (sample mean = $57.08, S= 10.52), 18 customers purchased dessert.

a) construct a 98% confidence interval estimate of the population mean amount spent per customer in the restaurant.

b) construct a 85% confidence interval estimate of the population proportion of customers who purchase dessert.

Answer 1

a)

One-Sample t-test Confidence Interval

The provided sample mean is Xˉ=57.08 and the sample standard deviation is s=10.52. The size of the sample is n = 60 and the required confidence level is 98%. Degree of freedom The number of degrees of freedom are df = 60 - 1 = 59, and the significance level is α=0.02. Critical Value Based on the provided information, the critical t-value for α=0.02 and df=59 degrees of freedom is tc​=2.3912. Margin of Error Therefore, based on the information provided, the 98% confidence for the population mean μ is calculated as: Therefore, the 98% confidence interval for the population mean μ is 53.8324<μ<60.3276, which indicates that we are 98% confident that the true population proportion μ is contained by the interval (53.8324,60.3276)

b)

One-Proportion Confidence Interval

We need to construct the 85% confidence interval for the population proportion. We have been provided with the following information: The sample size is N = 60, the number of favorable cases is X = 18 and the sample proportion is pˉ​=X/N​=18/60​=0.3, and the significance level is α=0.15 Critical Value Based on the information provided, the significance level is α=0.15, therefore the critical value is Zc​=1.4395. This can be found by either using excel or the Z distribution table. Margin of Error The confidence interval: Therefore, based on the data provided, the 85% confidence interval for the population proportion is 0.2148<p<0.3852, which indicates that we are 85% confident that the true population proportion p is contained by the interval (0.2148,0.3852)

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