A survey of 25 young professionals fond that they spend an average of $28 when dining...

A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a) Find and interpret the 95% confidence interval for the average spending. (b) Based on your answer in part (a), can you conclude statistically that the population mean is less than $30? Explain.

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Expert Solution

Solution :

Given that,

= $28

s =$10

n =25

Degrees of freedom = df = n - 1 =25 - 1 = 24

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2= 0.05 / 2 = 0.025

t /2,df = t0.025,24 = 2.064 ( using student t table)

Margin of error = E = t/2,df * (s / $\sqrt$ n)

= 2.064* ( 10/ $\sqrt$ 25)

= 4.1280

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

- E < < + E

28 - 4.1280 < <28 + 4.1280

23.8720 < < 32.1280

( 23.8720 , 32.1280 )

population mean is less than $30 , no less than is between 24 to 32

orchestra answered 2 years ago

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