Question

Assume that a sample is used to estimate a population mean μ. Find the 99.5% confidence interval for a sample of size 74 with a mean of 56.3 and a standard deviation of 19.1. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.

99.5% C.I. =

In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $33.9 and standard deviation of $6.1. Estimate how much a typical parent would spend on their child's birthday gift (use a 98% confidence level). Give your answers to 3 decimal places.

Express your answer in the format of ¯x ±  E.
$ ± $

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 56.3

sample standard deviation = s = 19.1

sample size = n = 74

Degrees of freedom = df = n - 1 = 74-1 = 73

t _/2,df = 2.807

Margin of error = E = t_/2,df * (s /n)

= 2.807 * (19.1 / 74)

Margin of error = E = 6.232

The 95% confidence interval estimate of the population mean is,

- E < <  + E

56.3 - 6.232 < < 56.3 + 6.232

50.067 < < 62.532

(50.067,62.532)

Point estimate = sample mean = =33.9

sample standard deviation = s = 6.1

sample size = n = 27

Degrees of freedom = df = n - 1 = 27 -1 =26

At 98% confidence level

= 1-0.98% =1-0.98 =0.02

/2 =0.02/ 2= 0.01

t/2,df = t_0.01,26

t/2,df = 2.479

The 98% confidence interval estimate of the population mean is,

  t_/2,df * (s /n)

33.9   2.479 *( 6.1 /27 )

33.9 2.910

($30.989 ,$ 36.810)