Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 randomly selected employees has a mean age of 22.4 years with a standard deviation of 3.8 years. Round to the nearest tenth. A) Determine the critical value ta/2 with n-the 1 degrees of freedom
B) Determine the lower and upper bound of the confidence interval
C) Interpret the confidence interval.

Confidence interval for mean, when population standard deviation is unknown:

, where = sample mean

n= sample size

s= sample standard deviation

= Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 ,

A) Critical t value for and degree of 18 freedom

= 2.8784

B) Required confidence interval:

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

marco saved more.   he saved $35 more.

step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35

savings account is an account only used for saving money. this acts as an emergency source of funding for an individual. answer a would result in the money in his account to grow as it specifically states that interest is compounded daily at the rate of 2%

step-by-step explanation: