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In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $50 and standard deviation of $7. Construct a confidence interval at a 99% confidence level.

Respuesta :

Answer:  99% confidence interval would be (45.74,54.26).

Step-by-step explanation:

Since we have given that

N = 18

Mean = $50

Standard deviation = $7

At 99% confidence level, z = 2.58

So, Interval would be

[tex]\bar{x}\pm z\dfrac{\sigma}{\sqrt{n}}\\\\=50\pm 2.58\times \dfrac{7}{\sqrt{18}}\\\\=50\pm 4.26\\\\=(50-4.26,50+4.26)\\\\=(45.74,54.26)[/tex]

Hence, 99% confidence interval would be (45.74,54.26).

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