A Fiji Travel Data Center survey reported that Fijians stayed an average of 7.5 nights when they
went on vacation around the country. For the purpose of the survey, the sample size was 1500.
Assume the population standard deviation was 0.8; find a point estimate of the population mean.
In addition, find the 95% confidence interval of the true mean.
Answers
A Fiji Travel Data Center survey reported that Fijians stayed an average of 7.5 nights when they
went on vacation around the country. For the purpose of the survey, the sample size was 1500.
Assume the population standard deviation was 0.8; find a point estimate of the population mean.
In addition, find the 95% confidence interval of the true mean.
Step-by-step explanation:
The sample has the following parameters
n=1500
µ =7.5 nights
σ=0.8 night ==>
The population has the following parameters
µ₀=µ=7.5 nights
σ₀=σ/sqrt(n)=0.8/sqrt(1500)
σ₀=0.0206559 night
The 95% confidence interval of the standardized normal distribution will found between the values of z1 and z2, ∴
z2= -z1 and
F(z2) - F(z1)=95%=0.95
F(z2) - F(z1)=1-2F(z1)=0.95 ==>
F(z1)=(1-0.95)/2
F(z1)=0.025 ==>
From the Standard normal cumulative probability table we find
z1= -1.96 ==>
z2= 1.96
The 95% confidence interval of the true mean ∴ will be
µ₀ + z1*σ₀<µ₀<µ₀+z2*σ₀
7.5 -1.96*0.0206559<µ₀<7.5+1.96*0.0206559
or
7.46 nights<µ₀<7.54 nights
or
7.46 nights; 7.54 nights