A research firm conducted a survey to determine the mean amount of money smokers spend on cigarettes during a day. A sample of 100 smokers revealed that the sample mean is N$ 5.24 and sample standard deviation is N$ 2.18 Assume that the sample was drawn from a normal population.
Answers
2.1. The sample mean $5.24 is a point estimate of the unknown population mean and is the best estimate, given the data.
2.2. The number of degrees of freedom are df=100-1=99,df=100−1=99, and the significance level is \alpha=0.05.α=0.05. The critical t-value for \alpha=0.05α=0.05 and df=99df=99 degrees of freedom is t_c=1.984217.t
c
=1.984217.
The 95% confidence for the population mean \muμ is computed using the following expression
CI=(\bar{x}-\dfrac{t_c\times s}{\sqrt{n}}, \bar{x}+\dfrac{t_c\times s}{\sqrt{n}})=CI=(
x
ˉ
−
n
t
c
×s
,
x
ˉ
+
n
t
c
×s
)=
=(5.24-\dfrac{1.984217\times 2.18}{\sqrt{100}}, 5.24+\dfrac{1.984217\times 2.18}{\sqrt{100}})==(5.24−
100
1.984217×2.18
,5.24+
100
1.984217×2.18
)=
=(4.807, 5.673)=(4.807,5.673)
Lower limit is 4.807.
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