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A survey of 50 retail stores revealed that the average price of a microwave was $375, with a sample standard deviation of $20. Assuming the population is normally distributed, what is the 95% confidence interval to estimate the true cost of the microwave?

Respuesta :

Answer:

(369.4563 , 380.5437)

Explanation:

Given that,

z_c = 1.96 at 95% confidence ⇒ Z- critical value

n = 50

x - bar = 375  

SD = 20

Margin of error, E = ?

[tex]E=\frac{z_c\times SD}{\sqrt{n} }[/tex]

[tex]E=\frac{1.96\times 20}{\sqrt{50} }[/tex]

E = 5.5437

Therefore,

95% CI:

= (375 - 5.5437, 375 + 5.5437)

= (369.4563 , 380.5437)

Hence, the 95% confidence interval to estimate the true cost of the microwave is (369.4563 , 380.5437).