A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 119 students who averaged 34.2 texts per day. The standard deviation was 12.7 texts. Round answers to 3 decimal places where possible.
a. To compute the confidence interval use a ? t z distribution.
b. With 98% confidence the population mean number of texts per day is between and texts.
c. If many groups of 119 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day.

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09-12-2022
Mathematics

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A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 119 students who averaged 34.2 texts per day. The standard deviation was 12.7 texts. Round answers to 3 decimal places where possible.
a. To compute the confidence interval use a ? t z distribution.
b. With 98% confidence the population mean number of texts per day is between and texts.
c. If many groups of 119 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day.

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varshamittal029 varshamittal029 · Answer 1 · 2022-12-18T11:30:57-05:00

a. t distribution

b. 98% confidence the population mean number of texts per day is between and texts is 31. 455 and 36.945

c. Two percent will not contain the true population mean number of texts per day.

Confidence interval

A confidence interval, in statistics, refers to the probability that a population parameter will fall between a set of values for a certain proportion of times.
Analysts often use confidence intervals than contain either 95% or 99% of expected observations.

Given that,

xbar = 34.2

s = 12.7

n = 119

degrees of freedom, df = n - 1 = 118

a. t distribution

b. CI level is 98%, hence α = 1 - 0.98 = 0.02

α/2 = 0.02/2

= 0.01,

tc = t (α/2, df) = 2.358

CI = [tex](34.2 - 2.358 * \frac{12.7}{\sqrt{119} } , 34.2 + 2.358 *\frac{12.7}{\sqrt{119} } )[/tex]

CI = (31.455 , 36.945)

With 98% confidence the population mean number of texts per day is between 31.455 and 36.945 texts.

(c) Approximately 98 percent of these confidence intervals will contain the true population mean number of texts per day, while approximately 2 percent will not.

To learn more about confidence interval check the given link

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