In a survey, 13 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $44 and standard deviation of $8. Use the theory-based inference applet to find the confidence interval at a 80% confidence level. Give your answers to one decimal place.

• We are 80% confident that the average (mean) amount that people spent on their child's last birthday gift, μ, was between  **** dollars and   **** dollars.
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• The sample statistic is  ****  dollars. 
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• The margin of error is  ****  dollars.
• The confidence interval can be written as ****  dollars  ±  **** dollars.

 

We are 80% confident that the average (mean) amount that people spent on their child's last birthday gift, μ, was between 40.99 dollars and 47.01 dollars.

The sample statistic is  1.356 dollars.

Margin of error = $ 3.01 dollars

The confidence interval can be written as 44 dollars ±3.01  dollars.

Step-by-step explanation

Point estimate = sample mean , x?  = $ 44 dollars

sample standard deviation , s = $ 8 dollars

sample size , n = 13

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 80% confidence level

Significance level, α = 1 - 80%

  =1 - 0.80 =0.20

α/2 = 0.10

t _α/2,df = t_0.1,12 = 1.356

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

= 1.356 * ( 8 / √ 13)

Margin of error , E = $ 3.0087 dollars

=$3.01

The 80% confidence interval estimate of the population mean

  CI= x? ± E  

= $ 44   ± $ 3.01

= ( $ 40.99, $ 47.01 )

We are 80% confident that the average (mean) amount that people spent on their child's last birthday gift, μ, was between 40.99 dollars and 47.01 dollars.