Question

Spring break can be a very expensive holiday. A sample of 95 students is surveyed, and the average amount spent by students on travel and beverages is $593.84. The sample standard deviation is approximately $369.34.

1. Construct a 90% confidence interval for the population mean amount of money spent by spring breakers.

Answer 1

Answer:

Given that,

= 593.84

Sample standard deviation = s = 369.34

Sample size=n= 95

Where is unknown so we are considering t - distribution

Degree of freedom = df= n-1 = 95-1= 94

90% confidence interval

c = 90%=90/100=0.90

= 1-c = 1-0.90=0.1

/2 = 0.1/2= 0.05

Critical value = t_c​​​= t_/2,df = t_0.05,94= 1.661

90% CI = t_/2,df *(s/sqrt(n))

90% CI = 593.84 1.661 *(369.34/sqrt(95))

90% CI = 593.84 62.941078

90% CI = (593.84-62.941078 , 593.84+62.941078)

90% CI = (530.898922 , 656.781078 )

90% CI = (530.899 , 656.781 )

90% confidence interval for the population mean amount of money spent by spring breakers is

530.899 < < 656.781

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