Question

In: Statistics and Probability

Problem one Suppose 34 households in my town were surveyed about the number of rolls of...

Problem one

Suppose 34 households in my town were surveyed about the number of rolls of toilet paper they had on Saturday, April 4. The mean number was found to be 6.5, and the standard deviation was calculated to be 2.1. Find a 90% confidence interval for the mean number of rolls of toilet paper possessed by a household in my town.

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 6.5

Population standard deviation = =2.1

Sample size = n =34

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)

=1.645 * ( 2.1 / $\sqrt$ 34 )

= 0.59
At 90% confidence interval
is,

- E < < + E

6.5 - 0.59 < < 6.5 + 0.59

5.91< < 7.09

(5.91 , 7.09)

orchestra answered 2 years ago

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