From a random sample of 68 businesses, it is found that the mean time that employees spend on personal issues each week is 5.8 hours with a standard deviation of 0.35 hours. What is the 95% confidence interval for the amount of time spent on personal issues?

(6.4005, 4.925)

Step-by-step explanation: n = sample size = 68

σ = population standard deviation = 0.35 hours

x = sample mean = 5.8

We are to construct a 95% confidence level for population mean.

This is given below as

Confidence interval = x ± Zα/2 × σ/√n

Where Zα/2 = critical value for a 2 tailed with 5 percent level of significance (that's α=5%) = 1.96

Note that level of significance + confidence level = 100, hence a 95% confidence level will give a 5% level of significance

For the upper limit of the 95% confidence interval, we have that

x + Zα/2 × σ/√n

5.8 +1.96(0.38/√68)

5.8 + 1.96 (0.046)

5.8 + 0.0875

= 6.4005

For the lower limit of the 95% confidence interval, we have that

x - Zα/2 × σ/√n

5.8 - 1.96(0.38/√68)

5.8 - 1.96 (0.046)

5.8 - 0.0875

=4.925

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of

x= 3^y

step-by-step explanation: