Question

In 2018, as a way of commemorating the 10-year anniversary of the release of the first smartphone, an undergraduate student polled a random sample of 26 her peers in hopes of estimating the average time (in hours) such students spend on their smartphone each day, on average. Using the data she collected, she produced a 95% confidence interval for the true mean time college students spend on their phone each day: (5.035, 6.165)

1. What was the margin of error of this confidence interval?

2. What was the sample mean time (in hours) for the n = 26 students?

3. What was the sample standard deviation (in hours) for the n = 26 students? Submit your answer rounded to four decimals.

Show work please and thank you.

Answer 1

1.

Formula for Confidence Interval for Population mean when population Standard deviation is not known

Margin of error

i.e

Confidence Interval for Population mean when population Standard deviation is not known

Given,

95% confidence interval for the true mean time college students spend on their phone each day: (5.035, 6.165)

i.e

- margin of error = 5.035 -------------------------------(1)

+ margin of error = 6.165  -----------------------------(2)

(2) - (1) : [ + margin of error] - [ - margin of error] = 6.165 - 5.035

2 x margin of error = 1.13

margin of error = 1.13/2 = 0.565

margin of error = 0.565

2.

Sample mean time :

- margin of error = 5.035 -------------------------------(1)

+ margin of error = 6.165  -----------------------------(2)

(2) + (1) : [ + margin of error] + [ - margin of error] = 6.165 + 5.035

2 x = 11.2

= 11.2/2 = 5.6

= 5.6

3.

s: Sample standard deviation

From 2 margin of error = 0.565

given n = 26;n-1 =26-1 =25

for 95% confidence level = (100-95)/100 =0.05

/2 = 0.05/2 =0.025

t/2,n-1 = t0.025,25 = 2.0595

Sample standard deviation = 1.3989