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.
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 = tc= t/2,df = t0.05,94= 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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