A New York restaurant would like to estimate the average amount customers spend for dinner. A sample of size 49 customers resulted an average of $32.60. Assume the population standard deviation is $5.What is the point estimation for the mean of the population?At 90% level of confidence, what is the margin of error?What is the 90% confidence level for the mean of the population?

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windyyork windyyork · Answer 1 · 2019-12-10T05:53:09+01:00

Answer: Point of estimation = 32.60

Margin of error = 1.175.

95% confidence interval would be (81.95,84.45).

Step-by-step explanation:

Since we have given that

n = 49

Average = $32.60

Standard deviation = $5

We need to find:

1) Point of estimation for the mean of the population:

[tex]\mu=\$32.60[/tex]

2) At 90% level of confidence, z = 1.645

Margin of error would be

[tex]z\dfrac{\sigma}{\sqrt{n}}\\\\=1.645\times \dfrac{5}{\sqrt{49}}\\\\=1.645\times \dfrac{5}{7}\\\\=1.175[/tex]

So, Margin of error = 1.175.

3) At 90% level of confidence, interval would be

[tex]\bar{x}\pm 1.175\\\\=32.60\pm 1.175\\\\=(32.60-1.175,32.60+1.175)\\\\=(31.425,33.775)\\\\=(31.43,33.76)[/tex]

Hence, 90% confidence interval would be (31.43,33.76).