Question

In a survey, 15 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $42 and standard deviation of $5. Estimate how much a typical parent would spend on their child's birthday gift (use a 90% confidence level).

Give your answers to one decimal place. Provide the point estimate and margin or error.

Do you add ( ) when answering?

Answer 1

Solution :

Given that,

Point estimate (x̅)= $42

s = $5

n = $15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 90% confidence level the t is ,

α= 1 - 90% = 1 - 0.90 = 0.1

α/ 2 = 0.1 / 2 = 0.05

t_α/2,df = t_0.05,14 = 1.761

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

= 1.761* (5 / √15)

E = 2.273

The 90% confidence interval estimate of the population mean is,

x̅- E < μ < x̅ + E

42 - 2.273 < μ < 42 + 2.273

39.727< μ < 44.273

($39.7,$44.3)

A typical parent should spend on their child birthday gift from $39.7 to $44.3.

Answer 2

SOLUTION :

Sample size, n = 15 . It is less than 30, so t-distribution will apply.

For sample of size 15, degree of freedom = 15 - 1 = 14.

Confidence level = 90% = 0.90

So, alpha = 1 - 0.90 = 0.10 (on two tail basis)

Hence, t  for df 14, and alpha = 0.10 (two tails) = 1.761 

Hence,  for 90% CL, spending can vary by 1.761 value of t  on either side of mean

SD of population = 5 / sqrt(n) = 5/sqrt(15) = 1.291 

So, Margin / Error , E  at 90% CL :

= +/- t * SD of population

=+/-  1.761 * 1.291

= +/- 2.2735 

= +/- 2.27 ($) approx. (ANSWER).

Parents spend at 90% CL :

Lower limit = Mean - E = 42 - 2.27 = 39.73 ($) (ANSWER)

UPPER LIMIT = Mean + E  = 42 + 2.27 = 44.27 ($) (ANSWER).