Question

Suppose that the number of dollars spent per week on groceries are normally distributed with an unknown mean and standard deviation. A random sample of 18 grocery bills is taken and gives a sample mean of 103 dollars and a sample standard deviation of 10 dollars.

The margin of error for a 98% confidence interval estimate for the population mean using the Student's t-distribution is 6.05. Find a 98% confidence interval estimate for the population mean using the Student's t-distribution.

Round the final answers to two decimal places.

Answer 1

We are given that the sample size $\sf\:n = 18$, sample mean $\sf\:\bar{X} = 103$, sample standard deviation $\sf\:S = 10$, and margin of error $\sf\:E = 6.05$ for a 98% confidence interval estimate for the population mean.

Using the formula for the margin of error, we can find the value of the t-score that corresponds to a 98% confidence level with 17 degrees of freedom (since we have 18 samples):

$\sf\:t_{0.01/2, 17} \implies 2.898$

Now, we can use the formula for a confidence interval estimate to find the lower and upper bounds of the interval:

$\sf\:\bar{X} - E \implies 103 - 6.05 < \mu < \bar{X} + E \implies 103 + 6.05$

Substituting in the values we have:

$\sf\:103 - 6.05 < \mu < 103 + 6.05$

Simplifying:

$\sf\leadsto\:96.95 < \mu < 109.05$

Therefore, a 98% confidence interval estimate for the population mean is $\sf\: (96.95, 109.05)$ rounded to two decimal places.

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