A researcher is interested in finding a 95% confidence interval for the mean num

Question

A researcher is interested in finding a 95% confidence interval for the mean number of times per day that college students text. The study included 210 students who averaged 28 texts per day. The standard deviation was 21 texts.

Round your answers to two decimal places.

A. The sampling distribution follows a                        [ Select ] ["F", "normal", "T", "Chi-square"] distribution.

B. With 95% confidence the population mean number of texts per day of is between                        [ Select ] ["24.92", "25.79", "27.37", "25.14"] and    [ Select ] ["31.19", "31.20", "29.28", "30.86"] texts.

C. If many groups of 210 randomly selected students are studied, then a different confidence interval would be produced from each group. About [ Select ] ["5", "95", "1", "99"] percent of these confidence intervals will contain the true population mean number of texts per day and about    [ Select ] ["95", "99", "5", "1"] percent will not contain the true population mean number of texts per day.

Explanation / Answer

A. The sampling distribution follows a T distriubtion.

B. Standard error of the mean (se0) = / sqrt(n) = 21/ sqrt(210) = 1.45

95% confidence interval = xbar +- t209,0.05  * se0

= 28 +- 1.971 * 1.45

= (25.14, 30.86)

C . Here, 95% of these confidence intervals will contain the true population mean number of texts per day and about 5% will not contain the true population mean number of texts per day.

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