A fast food restaurant executive wishes to know how many fast food meals adults eat each week. They want to construct a 90% confidence interval for the mean and are assuming that the population standard deviation for the number of fast food meals consumed each week is 1.4. The study found that for a sample of 1271 adults the mean number of fast food meals consumed per week is 3. Construct the desired confidence interval. Round your answers to one decimal place.

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yogeshkumar49685 yogeshkumar49685 · Answer 1 · 2022-07-16T04:02:25+02:00

The confidence interval which shows 90% confidence level for the mean and assumed standard deviation is (2.94 meals, 3.0565 meals).

Given standard deviation=1.4. Sample size =1271, mean=3.

We have to find the confidence interval for 90% confidence level.

We have to find out α level that is the subtraction of 1 by the confidence interval divided by 2.

α=(1-0.90)/2

=0.05

Now we have to find z in the z table as such z has a p value of 1-α so it is z with p value of 1-0.05=0.95

Z=1.44 from z table.

Now find M as such that

M=z* st/[tex]\sqrt{n}[/tex]

where st is standard deviation ,n is sample size.

M=1.44*1.4/[tex]\sqrt{1271}[/tex]

=1.44*1.4/35.65

=2.016/35.65

=0.0565

Lower end= mean -M

=3-0.0565

=2.94

Upper end=Mean+ M

=3+0.0565

=3.0565

Hence the confidence interval showing 90% confidence level is (2.94,3.0565).

Learn more about z test at https://brainly.com/question/14453510

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