Question

Prompt for question 1 part a-c: Smart phones have changed the world and how we spend our time. A group of researchers at BYU want to estimate the mean daily number of minutes that BYU students spend on their phones. In fall 2019, they took a random sample 430 BYU students and found that on average, students spend 312 minutes on their phones with a standard deviation of 54 minutes per day. The plot of the sample data showed no extreme skewness or outliers. Calculate a 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019. Do not round off in the intermediate steps, only the final answer should be rounded off to two decimal places.

STATE Calculate a 96% confidence interval estimate for the mean daily number of minutes that BYU students spend on their phones in fall 2019.

confidence level comes out to be: 96%

the t* critical value is 2.060012.

the margin of error is 5.36.

part a) How would selecting a 90% level of confidence change the size of the calculated confidence interval? . Explain or justify your answer by recalculating, rounded to two decimal places

part b) At a 90% level of confidence, what sample size would be needed to estimate the parameter of interest to within a margin of error of ± 10 minutes? Use σ = 56 minutes.

part c)Suppose that a second random sample of BYU students was conducted in fall 2019. Using this data, the 96% confidence interval was calculated to be (310.92, 317.58). Rounded to two decimal places, what is the margin of error for this confidence interval? Show your work.

Answer 1

l:

alpha=1-0.96=0.04

alpha/2=0,04/2=0.02

degrees of freedom=n-1=430-1=429

n=430

t crit==T.INV(0.02,429)=2.0600124

96% confidence interval for mean daily number of minutes that BYU students spend on their phones in fall 2019

xbar-Tc*s/sqrt/(n),xbar+Tc*s/sqrt(n)

312-2.0600124*54/sqrt(430),312+2.0600124*54/sqrt(430)

306.6355,317.3645

96% lower limit mean= 306.64

96% upper limit mean=317.36

Solution-2:

part a) How would selecting a 90% level of confidence change the size of the calculated confidence interval? . Explain or justify your answer by recalculating, rounded to two decimal place

alpha=1-0.90=0.10

alpha/2=0.10/2=0.05

t crit==T.INV(0.05,429)=1.648413266

90% confidence interval for mean daily number of minutes that BYU students spend on their phones in fall 2019

xbar-Tc*s/sqrt/(n),xbar+Tc*s/sqrt(n)

312-1.648413266*54/sqrt(430),312+1.648413266*54/sqrt(430)

307.7073,316.2927

90% lower limit mean= 307.71

90% upper limit mean=316.29

Solution-c

sample size=n=(Z*sigma/E)^2

=(1.645*56/10)^2

= 84.86094

=85

n=85

Solution-d:

MOE=margin of error

sample mean-MOE=310.92----(1)

sample mean+MOE=317.58-----(2)

add 2 eq

2*sample mean=310.92+317.58

=314.25

substitue in eq(2)

MOE=317.58-314.25=3.33

margin of error=3.33