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13#A toy manufacturer wants to see how long, on average, a new toy captures children's attention. He tests 11 children selected at random and finds that their mean attention span is 22 minutes with a standard deviation of 5 minutes. If we assume that attention spans are normally distributed, find a 90% confidence interval for the mean attention span of children playing with this new toy. Give the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas. )Lower limit:Upper limit:



Answer :

The lower limit of the confidence interval is 18.80 and the upper limit of the confidence interval is 25.19

Confidence interval = sample mean ± margin of error

Margin of error = z*population standard deviation/[tex]\sqrt{n}[/tex]

The level of confidence is represented by z==2.12

the size of the sample n=11

sample mean= 22

confidence interval = sample mean ± z*population standard deviation/[tex]\sqrt{n}[/tex]

upper limit of confidence interval = sample mean + z*population standard deviation/[tex]\sqrt{n}[/tex]

=22+2.12*5/[tex]\sqrt{11}[/tex]=25.19

lower limit of confidence interval = sample mean - z*population standard deviation/[tex]\sqrt{n}[/tex]

=22-2.12*5/[tex]\sqrt{11}[/tex]=18.80

Therefore, the lower limit of the confidence interval is 18.80 and the upper limit of the confidence interval is 25.19

To know more about confidence intervals:

https://brainly.com/question/24131141

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